Category: CBSE Class 10

Here we are providing Class 10 History Chapter 4 Extra Questions and Answers The Making of Global World was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-social-science/

The Making of Global World Class 10 Extra Questions History Chapter 4

Question 1.
What is silk route?
Answer:
It is a route through which the Chinese silk would move from China to West-bound regions.

Question 2.
Give examples of goods sent to Europe from Asia.
Answer:
Silk, pottery, textiles, spices.

Question 3.
What would Europe give to Asia in return to silk, spices etc.?
Answer:
Precious metals: gold and silver.

Question 4.
Who were the American Indians?
Answer:
America’s original inhabitants were known as the American Indians.

Question 5.
How do you say that a crop could make the difference between life and death? Give example.
Answer:
It a crop was available, people would eat it and live; if it was not available, people would die Europe’s poor lived as potato was available: the people of Ireland died when the potato crop was destroyed by disease.

Question 6.
When do you find world shrinking?
Answer:
The pre-modern world began shrinking when the European sailors found new sea route to Asia and discovered”Americas.

Question 7.
Name the countries where precious metals such as silver and gold were found?
Answer:
Present Peru and Mexico, and El Dorado.

Question 8.
Mention the problems Europe had faced until the 19th century.
Answer:
Poverty, hunger, diseases, religious confiets.

Question 9.
What do you mean by ‘flow of trade’?
Answer:
Flow of trade refers to trade in goods (cloth, wheat).

Question 10.
What do you mean by ‘flow of labour?
Answer:
Flow of labour refers to migration: people moving out in search of employment.

Question 11.
What do you mean by ‘flow of capital’?
Answer:
Flow of capital refers to investment of capital over long distances.

Question 12.
What were the com laws? Why were these laws later scrapped?
Answer:
Corn laws were laws enacted by Britain so to disallow ‘corns’ in the country. As food prices arose, the industrialists and the urban dwellers forced the government to withdraw these laws.

Question 13.
Why were certain colonies in Punjab described as the canal colonies?
Answer:
The British government built a network of canals, in the then semi deserts areas of Punjab. Colonies established around these canals were called the canal colonies.

Question 14.
Mention as to what had helped imagine a transformed 194-century world?
Answer:
Railways, steamship, telegraph etc. etc.

Question 15.
Trade and profit apart, what was the darker side of colonialism for the colonial people?
Answer:
Loss of their freedoms, and also of livelihood. The colonialists made use of the natural resources o the colonies for’ their colonial design.

Question 16.
Why were the Europeans attracted to Africa?
Answer:
The Europeans were attracted to Africa due to its vast resources of land and minerals, in late 19th century.

Question 17.
Why were the inheritance laws introduced in Africa?
Answer:
The purpose behind the inheritance laws was to displace the peasants from their ‘and, and force them to work as wages.

Question 18.
What was rinderpest?
Answer:
Rinderpest was a deadly cattle disease. It spread in Africa in 1890s.

Question 19.
From which regions of India did the indentured workers come to work with the plantation employers?
Answer:
Eastern Uttar Pradesh, Bihar, Madhya Pradesh, some districts of Tamil Nadu.

Question 20.
What is described as the new system of slavery?
Answer:
The 19th century indenture was described as the new system of slavery.

Question 21.
What is Hosay’?
Answer:
‘Hosay’ (for Imam Hussain) a notorious I carnival being transformed from Muharram.

Question 22.
Who were the Allies and the central powers during the World War I?
Answer:
Allies: Britain, France, Russia, later the USA.
Central Powers: Germany. Austria-Hungary j and the Ottoman Turkey.

Question 23.
Why do we say that the World War I | was the first modern industrial war?
Answer:
The World War I is described as the first modem industrial war because it saw the use of machine guns, tanks, aircrafts, chemical weapons etc. on a massive scale.

Question 24.
Why was Britain burdened with huge external debt after the World War I?
Answer:
Britain had borrowed liberally from the USA. It was unable to recover from economic losses during the war.

Question 25.
The World War I led to an economic boom initially and economic depression later. Explain.
Answer:
The World War I resulted in economic recovery initially leading to mass production as well as mass consumption in early 1920s. But soon mass production had no buyers, hence economic depression in 1929-30.

Question 26.
What was the effect the Great Depression on India?
Answer:
Foodgrains prices fell 50%; jute prices fell 60% all this leading to rural indebtedness.

Question 27.
At which conference was the International Monetary Fund established and for what purposes?
Answer:
At Bretton Woods conference, to deal with the external surpluses and deficits of its members.

Question 28.
Why was the World Bank established?
Answer:
The World Bank (earlier called the International Bank for Reconstruction and Development) was set up, after the World War II, to finance post-war reconstruction.

Question 29.
How have the human societies become interlinked? Explain.
Answer:
All through history, human societies have become steadily more interlinked. From ancient times, travellers, traders, priests and pilgrims travelled vast distances for knowledge, opportunity and spiritual fulfilment, or to escape persecution. They carried goods, money, values, skills, ideas, inventions, and even germs and diseases
Some such examples are:

• As early as 3000 BCE an active coastal trade linked the Indus valley civilisations with present- day West Asia,
• For more than a millennia, cowries (the Hindi coudi Seashells used as a form of currency) from the Maldives found their way to China and East Africa.
• The long distance spread of disease carrying germs may be traced as far back as the seventh century.

Question 30.
Give examples as to how food offers a link between peoples of different region.
Answer:
Food offers many examples of long distance cultural exchange Traders and travellers introduced new crops to the lands they travelled. Even ‘ready’ foodstuff in distant parts of the world might share common origins. Take spaghetti and noodles. It is believed that noodles travelled west from China to become spaghetti. Or, perhaps Arabs traders took pasta to fifth-century Sicily, an island now in Italy.

Question 31.
Explain what we mean when we say that the world shrank in the 1500s.
Answer:
By shrinking of the world what we mean is that the people of the world have come closer. With the new sea route to Asia, it was now approachable to the Europeans. The discovery of America, unknown until then, came to be known to the world. Europe, Asia, Europe, Australia – all these seemed nearer to each other.

Question 32.
How did smallpox pave way for the conquest of Americas by the Spanish.
Answer:
In fact, the most powerful weapon of the Spanish conqueror was not a conventional military weapon at all. It was the germs such as those of smallpox that they carried on their person. Because of their long isolation, America’s original inhabitants had no immunity against these diseases that can from Europe. Smallpox in particular provec a deadly killer. Once introduced, it spread deep into the continent, ahead even of any Europeans reaching there. It killed and decimated whole communities, paving the way for conquest.

Question 33.
Why did some Europeans move from | Europe to America during most part of the 17th 1 century?
Answer:
Until the nineteenth century, poverty and hunger were common in Europe. Cities were crowded and deadly diseases were widespread. Religious conflicts were common, and religious dissenters were persecuted. Thousands therefore fled Europe for America, where by the eighteenth century, plantations worked by slaves captured in Africa. Nearly 5 crore people migrated from Europe to America and Australia during the 19th century.

Question 34.
How did the coming of the world economy lead to large migration of the people.
Answer:
Increase in agricultural products led to sale abroad Railways, were needed to link the agricultural regions to the ports, New harbours had to be built and old ones expanded to ship the new cargoes, People had to settle on the lands to bring them under cultivation. This meant building homes and settlements. All these activities in turn required capital and labour.

Capital flowed from financial centres such as Landon. The demand for labour in places where labour was in short supply – as in America and Australia – led to more migration. All over the world some 150 million are estimated to have left their homes, crossed oceans and vast distances over land in search of a better future.

Question 35.
Give briefly the process of colonization of the world during the 19 century.
Answer:
The European powers carved up the borders of the African countries during late 19th century. In 1885, the big European powers met in Berlin to complete the carving up of Africa between them Britain and France made vast additions to their overseas territories in the late nineteenth century. Belgium and Germany became new colonial powers. The US also became a colonial power in the late 1890s by taking over some colonies earlier held by Spain.

The geographical explorations did contribute to the colonization of Africa. Sir Henry Morton Stanley refers about Livingston’s exploration, in helping the conquest of Africa.

Question 36.
Give a brief account of the rinderpest, the devastating cattle disease.
Answer:
Rinderpest arrived in Africa in the late 1880s. It was carried by infected cattle imported from British Asia to feed the Italian soldiers invading Eritrea in East Africa. Entering Africa in the east, rinderpest moved west like forest fire’, reaching Africa’s Atlantic coast in 1892. It reached the Cape.

Along the way rinderpest killed 90 per cent of the cattle. The loss of cattle destroyed African livelihoods. Planters, mine owners and colonial governments now successfully monopolised what scarce cattle resources remained, to strengthen their power and to force Africans into the labour market. Control over the scarce resource of cattle”enabled European colonisers to conquer and subdue Africa.

Question 37.
Explain with examples how the Indian bankers and traders financed export agriculture.
Answer:
The Indian Bankers and traders, such as Shikarpuri Shroffs and Nattukottai Chettiars financed export agriculture in Central and Southeast Asia, using either their own funds or those borrowed from European banks. They had a sophisticated system to transfer money over large distances, and even developed indigenous forms of corporate organisation.

Indian traders and moneylenders also followed European colonisers into Africa. Hyderabadi Sindhi traders, however, ventured beyond European colonies.

Question 38.
Give example as to how the trade in Indian textile declined but increased in raw material ?
Answer:
With stiff competition in international market, export in textile declined. It declined from some 30% around 1800 to 15% by 1815, and by
1870, it came below 3%. But the export in raw- material steadily increased between 1812 and 1871, the share of raw-cotton exports rose from 5% to 35%.

Question 39.
How was the economic depression harmful for jute production in India ?
Answer:
The economic depression (1929-30) proved harmful for the Indian jute producers. The producers grow raw jute that was processed in factories for export in the form of gunny bags. But as gunny exports collapsed, the price of raw jute crashed more than 60 percent. Peasants who borrowed in the hope of better times or to increase output in the hope of higher incomes faced ever lower prices, and fell deeper and deeper into debt.

Question 40.
Why did the Bretton Woods institutions shift their attention towards developing countries
Answer:
The IMF and the World Bank (the two Bretton Woods institutions) were designed to meet the financial needs of the industrial countries. They were not equipped to cope with the challenge of poverty and lack of development in the former colonies. But as Europe and Japan rapidly rebuilt their economies, they grew less dependent on the IMI and the World Bank. Thus from the late 1950s, the Bretton Woods institutions began to shift their attention more towards developing countries.

Question 41.
Discuss the importance of language and popular traditions in the creation of national identity.
Answer:

• A nation is more important than an individual. So a person is known by his/her language he/she speaks or the traditions are follows. For example, if a person speaks Hindi in England and celebrates Holi or Diwali there, he/ she will be identified as an Indian easily.
• Since traditions and languages take a very long time to develop in a particular region, these get deeply rooted among the people of that region.
• People may come and people may go but languages and traditions go on forever, although these might change their form.

Question 42.
Briefly summarise the two lessons learnt by economists and politicians from the inter-war economic experience?
Answer:

1. Since the wars destroyed the economies of many countries of the world and pushed them back by many years so the economists, as well as the politicians, thought that they must work very hard for the economic development and stability of the industrialised countries.
2. Now, they understood the value of interdependence of different national economies all over the world, which slowly transformed into the era of globalisation.

Question 43.
Substantiate with examples how the silk routes were helpful in linking the world?
Answer:
The silk routes are a good example of pre-modern trade and cultural links between distant parts of the world. The name silk routes points to the importance of West-bound Chinese silk cargoes along this route. Historians have identified several silk routes, overland and by sea, knitting together vast regions of Asia, and linking Asia with Europe and northern Africa.

They are known to have existed since before the Christian Era and thrived almost till the fifteenth century. But Chinese pottery also travelled the same route, as did textiles and spices from India and Southeast Asia. In return, precious metals – gold and silver – flowed from Europe to Asia.

Trade and cultural exchange always went hand in hand. Early Christian missionaries almost certainly travelled this route to Asia, as did early Muslim preachers a few centuries later. Much before all this: Buddhism emerged from eastern India and spread in several directions through intersecting points on the silk routes.

Question 44.
Do you agree that by 1890, a global agricultural economy had taken place?
Answer:
By 1890, a global agricultural economy had taken shape, accompanied by complex changes in labour movement patterns, capital flows, ecologies and technology. Foud no longer came from a nearby village or town, but from thousands of miles away. It was not grown by a peasant tilling his own land, but by an agricultural worker, perhaps recently arrived, who was now working on a large farm that only a generation ago had most likely been a forest.

It was transported by railway, built for that very purpose, and by MPs which were increasingly manned in these decades by low-paid workers from southern Europe, Asia, Africa and the Caribbean.

Indeed, so rapidly did regional specialisation in the production of commodities develop, that between 1820 and 1914 world trade is estimated to have multiplied 25 to 40 times. Nearly 60 per cent of this trade comprised ‘primary products’ that is agricultural products such as wheat and cotton, and minerals such as coal.

Question 45.
How did the Great Depression affect the US economy?
Answer:
The US was also the industrial country most severely affected by the depression. With the fall in prices and the prospect of a depression, US banks had also slashed domestic lending and called back loans, Farms could not sell their harvests, households were ruined, and businesses collapsed. Faced with falling incomes.

What are MNCs?
Multinational corporations (MNCs) are large companies that operate in several countries at the same time. The first MNCs were established in the 1920s Many more came up in the 1950s and 1960s as US businesses expanded worldwide and Western Europe and Japan also recovered to become powerful industrial economies.

The worldwide spread of MNCs was a notable feature of the 1950s and 1960s. This was partly because high import tariffs imposed by different governments forced MNCs to locate their manufacturing operations and become ‘domestic producers’ in as many countries as possible.

Many households in the US could not repay what they had borrowed and were forced to give up their homes, cars and other consumer durables. The consumer’s prosperity of the 1920s now disappeared in a puff of dust. As unemployment soared, people trudged long distances looking for any work they could find.

Ultimately, the US banking system itself collapsed. Unable to recover investments, collect loans and repay depositors.
thousands of banks went bankrupt and were forced to close. The numbers are phenomenal. by 1933 over 4,000 banks had closed and between 1929 and 1932 about 110,000 companies had collapsed.

Question 46.
How would you explain that from mid-1970s, international financial system substantially changed?
Answer:
From the mid-1970s the international financial system also changed in important ways.

(i) Earlier, developing countries could turn to International institutions for loans and development assistance. But now they were
forced to borrow from Western commercial banks and private lending institutions. This led to periodic debt crises in the developing world, and lower incomes and increased poverty, especially in Africa and Latin America.

(ii) The índustrial world was also hit by unemployment that began rising from the mid 1970s and remained high until the early 1990s. From the late 1970s MNCs also began to shift production operations to low wage Asian countries.

(iii) China had been cut off from the post war world economy since its revolution in 1949. But new economic policies in China and the collapse of the Soviet Union and Soviet-style communism in Eastern Europe brought many countries back into the fold of the world economy.

Map Skill

On the outline map of Africa, draw map of colonial Africa at the end of the 19th century.
Answer:
See map below:

Objective Type Questions

1. Fill in the blanks with the appropriate words.

(i) …………………………….. or seashells were used as a form of currency.
Answer:
Cowdies

(ii) Most of our common foods came from ……………………………… .
Answer:
American Indians,

(iii) Silk routes pertain to a particular country called …………………………….. .
Answer:
China

(iv) The Great Depression occurred during …………………………….. .
Answer:
1929-30.

(v) Most India …………………………….. workers come from eastern Uttar Pradesh, Bihar, Madhya Pradesh.
Answer:
indenture.

2. Choose the most appropriate alternative:

(i) ‘Chutney motive was popular in
(a) Chicago
(b) Landon
(c) Trinidad
(d) Tokyo
Answer:
(c) Trinidad

(ii) The headquarter of the English Last India company was in:
(a) Yorkshire
(b) Landon
(c) Glasgow
(d) Liverpool
Answer:
(b) Landon

(iii) One of the Following was the producer of car in the USA
(a) Henry Ford
(b) Farty Taft
(c) Gerald Ford
(d) Bill Gates
Answer:
(a) Henry Ford

(iv) Group of 77 belong to the following nations:
(a) Developed
(b) Developing
(c) African
(d) Asian
Answer:
(b) Developing

(v) One of the following in Bretton Woods institutions
(a) IMF
(b) WHO
(c) ILO
(d) UNESCO
Answer:
(a) IMF

Extra Questions for Class 10 Social Science

Here we are providing Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Polynomials with Answers Solutions

Extra Questions for Class 10 Maths Chapter 2 Polynomials with Solutions Answers

Polynomials Class 10 Extra Questions Very Short Answer Type

The graphs of y = p(x) for some polynomials (for questions 1 to 4) are given below. Find the number of zeros in each case.

Polynomials Class 10 Extra Questions Question 1.

Answer:
There is no zero as the graph does not intersect the X-axis.

Polynomial Class 10 Extra Questions Question 2.

Answer:
The number of zeros is four as the graph intersects the X-axis at four points.

Polynomials Class 10 Extra Questions With Answers Question 3.

Answer:
The number of zeros is three as the graph intersects the X-axis at three points.

Class 10 Polynomials Extra Questions Question 4.

Answer:
The number of zeros is three as the graph intersects the X-axis at three points.

Class 10 Maths Chapter 2 Extra Questions Question 5.
What will the quotient and remainder be on division of ax² + bx + c by px² + qx² + rx + 5, p ≠ 0?
Answer:
0, ax² + bx + C.

Polynomials Class 10 Questions With Answers Question 6.
If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p(x) and g(x)?
Answer:
Since the quotient is zero, therefore
deg p(x) < deg g(x)

Polynomials Extra Questions Class 10 Question 7.
Can x – 2 be the remainder on division of a polynomial p(x) by x + 3?
Answer:
No, as degree (x – 2) = degree (x + 3)

Extra Questions Of Polynomials Class 10 Question 8.
Find the quadratic polynomial whose zeros are -3 and 4.
[NCERT Exemplar]
Answer:
Sum of zeros = -3 + 4 = 1,
Product of zeros = – 3 x 4 = -12
∴ Required polynomial = x² – x – 12

Extra Questions On Polynomials Class 10 Question 9.
If one zero of the quadratic polynomial x² – 5x – 6 is 6 then find the other zero.
Answer:
Let α,6 be the zeros of given polynomial.
Then α + 6 = 5 3 ⇒ α = -1

Extra Questions For Class 10 Maths Chapter 2 Question 10.
If both the zeros of the quadratic polynomial ax² + bx + c are equal and opposite in sign, then find the value of b.
Answer:
Let α and -α be the roots of given polynomial.
Then α + (-α) = 0 ⇒ \(-\frac{b}{a}=0\) ⇒ b = 0.

Polynomial Extra Questions Class 10 Question 11.
What number should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the polynomial?
Answer:
Let f(x) = x² – 5x + 4
Then f(3) = 3² – 5 x 3 + 4 = -2
For f(3) = 0, 2 must be added to f(x).

Polynomials Extra Questions Question 12.
Can a quadratic polynomial x² + kx + k have equal zeros for some odd integer k > 1?
Answer:
No, for equal zeros, k = 0,4 ⇒ k should be even.

Class 10 Maths Polynomials Extra Questions Question 13.
If the zeros of a quadratic polynomial ax² + bx + c are both negative, then can we say a, b and c all have the same sign? Justify your answer.
Answer:
Yes, because \(-\frac{b}{a}\) = sum of zeros < 0, so that \(\frac{b}{a}=0\) > 0. Also the product of the zeros = \(\frac{c}{a}=0\) > 0.

Chapter 2 Maths Class 10 Extra Questions Question 14.
If the graph of a polynomial intersects the x-axis at only one point, can it be a quadratic polynomial?
Answer:
Yes, because every quadratic polynomial has at the most two zeros.

Extra Questions Of Polynomials Class 10 With Solutions Question 15.
If the graph of a polynomial intersects the x-axis at exactly two points, is it necessarily a quadratic polynomial?
Answer:
No, x⁴ – 1 is a polynomial intersecting the x-axis at exactly two points.

Polynomials Class 10 Extra Questions Short Answer Type 1

Question 1.
If one of the zeros of the quadratic polynomial f(x) = 4x² – 8kx – 9 is equal in magnitude but opposite in sign of the other, find the value of k.
Answer:
Let one root of the given polynomial be α.
Then the other root = -α
Sum of the roots = (-α) + α = 0
⇒ \(-\frac{b}{a}\) = 0 or \(-\frac{8k}{4}\) = 0 or k = 0

Question 2.
If one of the zeros of the quadratic polynomial (k – 1)x² + kx + 1 is -3 then find the value of k.
Answer:
Since – 3 is a zero of the given polynomial
∴ (k – 1)(-3)² + k(-3) + 1 = 0 :
⇒ 9k – 9 – 3k + 1 = 0 ⇒ k = 4/3.

Question 3.
If 1 is a zero of the polynomial p(x) = ax² – 3(a – 1)x -1, then find the value of a.
Answer:
Put x = 1 in p(x)
∴ p(1) = a(1)² – 3(a – 1) x 1 – 1 = 0
⇒ a – 3a + 3 – 1 = 0 ⇒ 2a = -2 ⇒ a = 1

Question 4.
If α and β are zeros of polynomial p(x) = x² – 5x + 6, then find the value of α + B – 3aß.
Answer:
Here, α + β = 5, αβ = 6
= α + β – 3αβ = 5 – 3 x 6 = -13

Question 5.
Find the zeros of the polynomial p(x) = 4x² – 12x + 9.
Answer:
p(x) = 4x² – 12x + 9 = (2x – 3)²
For zeros, p(x) = 0
⇒ (2x – 3)(2x – 3) = 0 ⇒ x = \(\frac{3}{2}, \frac{3}{2}\)

Question 6.
If one root of the polynomial p(y) = 5y² + 13y + m is reciprocal of other, then find the value of m.
Answer:

Question 7.
If α and β are zeros of p(x) = x² + x – 1, then find \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
Here, α + β = -1, αβ = -1,
So \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-1}=1\)

Question 8.
Given that one of the zeros of the cubic polynomial ax³ + bx² + cx + d is zero, find the product of the other two zeros.
Answer:
Let α, β, γ be the roots of the given polynomial and α = 0.
Then αβ + βγ + γα = c/a ⇒ βγ = c/a

Question 9.
If the product of two zeros of the polynomial p(x) = 2x³ + 6x² – 4x + 9 is 3, then find its third zero.
Answer:
Let α, β, γ be the roots of the given polynomial and αβ = 3
Then αβγ = \(-\frac{9}{2}\)
⇒ 3 x γ = \(\frac{-9}{2}\) or γ = \(\frac{-3}{2}\)

Question 10.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.
(i) \(-\frac{1}{4}, \frac{1}{4}\)
(ii) \(\sqrt{2}, \frac{1}{3}\)
Answer:
Let α, β be the zeros of polynomial.
(i) We have, α + β = \(-\frac{1}{4}\) and αβ = \(\frac{1}{4}\)
Thus, polynomial is
p(x) = x2 – (a + B) x + aß

Quadratic polynomial 4x² + x + 1

(ii) We have, α + β = √2 and  αβ = \(\frac{1}{3}\)
Thus, polynomial is p(x) = x² – (α + β) x + αβ
= x² – √2x + \(\frac{1}{3}\) = \(\frac{1}{3}\) (3x² – 3√2x + 1)
Quadratic polynomial = 3x² – 3√2x + 1

Polynomials Class 10 Extra Questions Short Answer Type 2

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients (Q. 1 – 2).

Question 1.
6x² – 3 – 7x
Answer:
We have, p(x) = 6x² – 3 – 7x
p(x) = 6x² – 7x – 3
(In general form)
= 6x² – 9x + 2x – 3 = 3x (2x – 3) + 1 (2x – 3)
= (2x – 3) (3x + 1)
The zeros of polynomial p(x) is given by
p(x) = 0) = (2x – 3) (3x + 1) = 0 ⇒ \(x=\frac{3}{2},-\frac{1}{3}\)
Thus, the zeros of 6x² – 7x – 3 are α = \(-\frac{3}{2}\) and β = \(-\frac{1}{3}\)
Now, sum of the zeros = α + β = \(\frac{3}{2}-\frac{1}{3}\) = \(\frac{9-2}{6}=\frac{7}{6}\)

Question 2.
4u² + 8u
Answer:
We have, p(u) = 4u² + 8u = p(u) = 4u (u + 2)
The zeros of polynomial p(u) is given by
p(u) = 0 ⇒ 4u (u + 2) = 0 .
∴ u = 0, -2
Thus, the zeros of 4u² + 8u are α = 0 and β = -2
Now, sum of the zeros = α + β = 0 – 2 = -2

Question 3.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2 (ii) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
Answer:
(i) We have,

Clearly, remainder is zero, so x2 + 3x + 1 is a factor of polynomial 3x⁴ + 5x³ – 7x² + 2x + 2

(ii) We have,

Clearly, remainder is zero, so t’ – 3 is a factor of polynomial 2t⁴ + 3t³ – 2t² – 9t – 12.

Question 4.
If α and β are the zeros of the quadratic polynomial f(x) = 2x² – 5x + 7, find a polynomial whose zeros are 2α + 3β and 3α + 2β.
Answer:
Since α and β are the zeros of the quadratic polynomial f(x) = 2x² – 5x + 7

Let S and P denote respectively the sum and product of the zeros of the required polynomial.

Question 5.
What must be subtracted from p(x) = 8x⁴ + 14x³ – 2x² + 7x – 8 so that the resulting polynomial
is exactly divisible by g(x) = 4x² + 3x – 2?
Answer:
Let y be subtracted from polynomial p(x)
: 8x⁴ + 14x³ – 2x² + 7x – 8 – y is exactly divisible by g(x)
Now,

∵ Remainder should be 0.
∴ 14x – 10 – y = 0 or 14x – 10 = y or y = 14x – 10
∴ (14x – 10) should be subtracted from p(x) so that it will be exactly divisible by g(x)

Question 6.
What must be added to f(x) = 4x⁴ + 2x³ – 2x² + x – 1 so that the resulting polynomial is divisible
by g(x) = x² + 2x – 3?
Answer:
By division algorithm, we have
f(x) = g(x) × q(x) + r(x)
= f(x) – r(x) = g(x) × q(x) ⇒ f(x) + {-r(x)} = g(x) × q(x)
Clearly, RHS is divisible by g(x). Therefore, LHS is also divisible by g(x). Thus, if we add –r(x) to f(x), then the resulting polynomial is divisible by g(x). Let us now find the remainder when f(x) is divided by g(x).

∴ r(x) = -61x + 65 or -r(x) = 61x – 65
Hence, we should add –r(x) = 61x – 65 to f(x) so that the resulting polynomial is divisible by g(x).

Question 7.
Obtain the zeros of quadratic polynomial 3x² – 8x + 4√3 and verify the relation between its zeros and coefficients.
Answer:
We have,

Question 8.
If α and β are the zeros of the polynomial 6y² – 7y + 2, find a quadratic polynomial whose zeros are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\)
Answer:
Let p(y) = 6y² – 7y + 2

Question 9.
If one zero of the polynomial 3x² – 8x + 2k + 1 is seven times the other, find the value of k.
Answer:
Let α and β be the zeros of the polynomial. Then as per question β = 7α

Question 10.
If one zero of the polynomial 2x² + 3x + λ is 1/2 find the value of and other zero.
Answer:
Let P(x) = 2x² + 3x + λ

Question 11.
If one zero of polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other, find the value of a.
Answer:
Let one zero of the given polynomial be α.
Then, the other zero is 1/α
∴ Product of zeros = α × \(\frac{1}{\alpha}\) = 1
But, as per the given polynomial product of zeros = \(\frac{6 a}{a^{2}+9}\)
∴ \(\frac{6 a}{a^{2}+9}\) = 1 ⇒ a² + 9 = 6a
⇒ a² – 6a + 9 = 0) ⇒ (a – 3)² = 0
⇒ a – 3 = 0 ⇒ a = 3
Hence, a = 3.

Question 12.
If the polynomial (x⁴ + 2x³ + 8x² + 12x + 18) is divided by another polynomial (x² + 5), the remainder comes out to be (px +q). Find values of p and q.
Answer:
Let f(x) = (x⁴ + 2x³ + 8x² + 12x + 18) and g(x) = (x² + 5)
On dividing f(x) by g(x), we get

Now, px + 9 = 2x + 3 ⇒ p = 2,q = 3 (By comparing the coefficient of x and constant term).

Polynomials Class 10 Extra Questions Long Answer Type 1

Question 1.
Verify that the numbers given alongside the cubic polynomial below are their zeros. Also verify the relationship between the zeros and the coefficients.
x³ – 4x² + 5x – 2; 2,1,1
Solution:
Let p(x) = x³ – 4 x² + 5x – 2
On comparing with general polynomial px) ax³ + bx² + cx + d, we get a = 1, b = -4, c = 5 and d = -2
Given zeros 2, 1, 1.
∴ p(2) = (2)³ – 4(2)² + 5(2) – 2 = 8 – 16 + 10 – 2 = 0
and p(1) = (1)³ – 4(1)² + 5(1) – 2 = 1 – 4 + 5 – 2 = 0
Hence, 2, 1 and I are the zeros of the given cubic polynomial.
Again, consider α = 2, β = 1, γ = 1
∴ α + 13 + y = 2 + 1 + 1 = 4

Question 2.
Find a cubic polynomial with the sum of the zeros, sum of the products of its zeros taken two at a time, and the product of its zeros as 2, -7, -14 respectively.
Solution:
Let the cubic polynomial be p(x) = ax³ + bx² + cx + d. Then

p(x) = a[x³ + (-2)x² + (-7)x + 14] ⇒ p(x) = a[x3 – 2x² – 7x + 14]
For real value of a = 1, p(x) = x³ – 2x² – 7x + 14

Question 3.
Find the zeros of the polynomial f(x) = x3 – 5x² – 2x + 24, if it is given that the product of its two zeros is 12.
Solution:
Let α, β and γ be the zeros of polynomial (fx) such that αβ = 12.

Now, α + β + γ = 5 α + β – 2 = 5
= α + β = 7 a = 7 – β
= (7 – β) β =12 ⇒ 7β – β² – 12
= β² + 7β + 12 = 0 ⇒ β² – 3β – 4β + 12 = O
= β = 4 or β = 3
β = 4 or β = 3
∴ α = 3 or α = 4

Question 4.
If the remainder on division of x3 – kx² + 13x – 21 by 2x – 1 is -21, find the quotient and the value of k. Hence, find the zeros of the cubic poIyncmia1 x³ – kx² + 13x.
Solution:
Let f(x) = x³ – kx² + 13x – 21

Question 5.
Obtain all other zeros of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeros are \(\sqrt{\frac{5}{3}}\) and \(\sqrt{\frac{5}{3}}\).
Solution:

Question 6.
Given that √2 is a zero of the cubic polynomial 6x³ + √2x² – 10x – 4√2, find its other zeros.
Solution:
The given polynomial is f(x) = (6x³ +√2x² – 10x – 4√2). Since √2 is the zero of f(x), it follows that (x – √2) is a factor of f(x).
On dividing f(x) by (x – √2), we get

Polynomials Class 10 Extra Questions HOTS

Question 1.
If α, β, γ bezerosofpo1ynomia1 6x³ + 3x² – 5x + 1, then find die value of α^-1 + β^-1 + γ^-1.
Solution:
∵ p(x) = 6x³ + 3x² – 5x + 1 so a = 6, b = 3, c = -5, d = 1
∴ α, β and γ are zeros of the polynomial p(x).

Question 2.
Find the zeros of the polynomial f(x) = – 12x² + 39x – 28, if it is given that the zeros are in AP.
Solution:
If α, β, γ are in AP., then,
β – α = γ + β ⇒ 2β = α + γ
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-(-12)}{1}\) = 12 ⇒ α + γ = 12 – β …….. (i)
From (i) and (ii)
2β = 12 – β or 3β = 12 or β = 4
Putting the value of β in (i), we have
8 = a + γ
αβγ = – \(\frac{d}{a}\) = \(\frac{-(-28)}{1}\) = 28 …….. (iii)
(αγ) 4 = 28 or αγ = 7 or γ = \(\frac{7}{α}\) ….. (iv)
Putting the value of γ = \(\frac{7}{α}\) in (iii), we get
⇒ 8 = α + \(\frac{7}{α}\) ⇒ 8α = α2 + 7
⇒ α² – 8α + 7 = 0 ⇒ α² – 7α – 1α + 7 = 0
⇒ α(α – 7)-1 (α – 7) = 0 ⇒ (α – 1)(α – 7) = 0
⇒ α = 1 or α = 7

Question 3.
If the polynomial f(x) = x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a. Find k and a.
Solution:
By division algorithm, we have Dividend = Divisor × Quotient + Remainder
⇒ Dividend – Remainder = Divisor × Quotient
⇒ Dividend – Remainder is always divisible by the divisor.
When f(x) = – 6x³ + 16x² – 25x + 10 is divided by x² – 2x + k the remainder comes out to be x + a.
∴ f(x) – (x + a) = x⁴ – 6x³+ 16x² – 25x + 10 – (x + a)
= x⁴ – 6x³ + 16x² – 25x + 10 – x – a x⁴ – 6x³ + 16x² – 26x + 10 – a
is exactly divisible by x² – 2x + k
Let us now divide x⁴ – 6x³ + 16x² – 26x + 10 – a by x² – 2x + k.

For f(x) – (x + a) = x⁴ – 6x³ + 16x² – 26x + 10 – a to be exactly divisible by x² – 2x + k, we must
have (-10 + 2k)x + (10 – a – 8k + k²) = 0 for all x
= – 10 + 2k = 0 and 10 – a – 8k + k² = 0
⇒ k = 5 and 10 – a – 40 + 25 = 0
⇒ k = 5 and a – 5

In this page, we are providing Sources of Energy Class 10 Extra Questions and Answers Science Chapter 14 pdf download. NCERT Extra Questions for Class 10 Science Chapter 14 Sources of Energy with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 14 Extra Questions and Answers Sources of Energy

Extra Questions for Class 10 Science Chapter 14 Sources of Energy with Answers Solutions

Extra Questions for Class 10 Science Chapter 14 Very Short Answer Type

Sources Of Energy Class 10 Extra Questions And Answers Question 1.
Define fossil fuel.
Answer:
Fossil fuels were formed millions of years ago, when plants and animal remains got buried under the earth and were subjected to high temperature and pressure conditions. For example, Coal, Petroleum, etc.

Sources Of Energy Class 10 Questions And Answers Question 2.
Write down the disadvantages of fossil fuels.
Answer:
These fossil fuels are non-renewable sources of energy and cause environmental problems due to pollution.

Sources Of Energy Class 10 Important Questions And Answers Question 3.
Define Nuclear energy.
Answer:
Nuclear energy: Energy released when some changes take place in the nucleus of the atom of a substance, is called Nuclear energy.

Questions On Sources Of Energy Class 10 Question 4.
Define solar cell.
Answer:
Solar cell is a device that converts solar energy into electricity.

Source Of Energy Class 10 Questions And Answers Question 5.
Define non-conventional sources of energy.
Answer:
Non-conventional sources of energy are those which are not used as the conventional ones and meet our energy requirement only on a limited scale.
Examples:

• Tidal energy
• Geothermal energy.

Sources Of Energy Extra Questions Question 6.
Define conventional sources of energy.
Answer:
Conventional sources of energy are those which are used extensively and meet a major portion of our energy requirement.
Examples:

• Fossil fuels
• Hydropower plants.

Sources Of Energy Questions And Answers Question 7.
How is charcoal produced and what is the advantage of charcoal in comparison to wood?
Answer:
When wood is burnt in a limited supply of oxygen, the volatile materials present in it get removed, and charcoal is produced. Charcoal bums without flames and is comparatively less smoky.

Extra Questions for Class 10 Science Chapter 14 Short Answer Type I

Extra Questions Of Chapter 14 Class 10 Science Question 1.
Write down the characteristics of a good fuel.
Answer:
Characteristics of a good fuel:

• Has high calorific value
• Produces less smoke
• Produces less residue after burning
• Easy availability
• Inexpensive
• Easy to store and transport

Class 10 Sources Of Energy Question Answer Question 2.
What is a thermal power plant?
Answer:
In a thermal power plant

• Coal, petroleum and natural gas is used to produce thermal electricity.
• Electricity transmission is very efficient.
• The steam produced by burning fossil fuels runs the turbine to produce electricity.

Source Of Energy Class 10 Important Questions Question 3.
Write down the advantages and disadvantages of a nuclear power plant.
Answer:
Advantage of Nuclear energy:

• Alternative source of energy due to depletion of fossil fuels.
• From a small amount of fuel, a large amount of energy is released.

Disadvantages of nuclear energy:

• Difficult to store and dispose nuclear waste which may cause environmental contamination.
• High cost of setting up a nuclear plant.
• Limited availability of raw material (uranium).

Ch 14 Science Class 10 Extra Questions Question 4.
Describe energy harnessed from wind and write down its advantages and disadvantages.
Answer:
Wind energy:

• It can converted into mechanical and electrical energy.
• Kinetic energy of the wind is used for running windmills, which can be used to lift water, grind grains, etc.

Advantages:

• Eco friendly
• Renewable

Disadvantages:

• Wind speed not uniform always.
• Needs a large area to erect series of windmills.
• Big amount of investment is needed.
• Output is less as compared to investment.

Extra Questions for Class 10 Science Chapter 14 Short Answer Type II

Sources Of Energy Class 10 Questions And Answers Pdf Question 1.
How can solar energy be harnessed? Mention any two limitations of using solar energy. Write down the advantages of solar cell.
Answer:
Solar energy:
Solar radiations can be converted into electricity through solar cells (photovoltaic cells). Photovoltaic cells convert solar radiations directly into electricity through silicon solar cells. Solar cells are arranged on large flat sheets to form a solar panel.

Advantages of solar cell:

• Solar cell have no moving parts, require little maintenance and work quite satisfactorily.
• They can be setup in remote and inaccessible hamlets or very sparsely inhabited areas.

Limitations:

• Solar cells are expensive.
• Solar devices are only useful during day time and on a sunny day.

Sources Of Energy Questions Question 2.
Write two different ways of harnessing energy from the ocean.
Answer:
Energy from the sea:
1. Tidal energy: Locations in India – Gulf of Kutch, Gujarat and West Bengal

• Depends upon harnessing the rise and fall of sea level due to tidal action.
• Dams are constructed across a narrow part of sea the and a turbine converts tidal energy into electrical energy.

Disadvantage: Uniform tidal action is not seen.

2. Wave energy:

• Kinetic energy of the sea waves are used to rotate turbines.
• These turbines generate electrical energy.

Question 3.
Write a short note on geothermal energy.
Answer:

• Energy harnessed from the heat of the Earth is called geothermal energy.
• Magma is formed when this heat melts the rocks. The molten rocks and hot gases are called magma.
• The magma gets collected at some depths below the earth’s surface. These places are called “hot spots’.
• When underground water comes in contact with these hot spots, it changes into steam, which can be used to generate electricity.

Advantages of geothermal energy:

• Renewable.
• Cost of production is not much.

Disadvantages of geothermal energy:

• Only few sites available for harnessing energy.
• Expensive to set up.

Extra Questions for Class 10 Science Chapter 14 Long Answer Type

Question 1.
What is biomass? Explain the principle and working of a biogas plant using a labelled schematic diagram. Write down its advantages.
Answer:
Biomass is a source of conventionally used fuels that are used in our country, e.g., cow dung cakes, fire-wood, coal, charcoal.

Biogas: It is a mixture of gases produced during the decomposition of biomass in the absence of oxygen.
Methane is the major component of biogas. (Biogas contains 75% methane, carbon dioxide, hydrogen and hydrogen sulfide).

Biogas plant: Animal dung, sewage, crop residues, vegetable wastes, poultry droppings, etc. are used to produce biogas in biogas plants.

Construction and working:
The plant has a dome-like structure built with bricks. A slurry of cow dung and water is made in the mixing tank from where it is fed into the digester. The digester is a sealed chamber in which there is no oxygen. Anaerobic microorganisms, that do not require oxygen, decompose or breakdown complex compound of cow dung slurry and produce methane, carbon dioxide, hydrogen and hydrogen sulfide.

Advantages of biogas

• It bums without smoke and leaves no residue.
• Its heating capacity is high.
• The slurry of biogas plant is used as an excellent manure, rich in nitrogen and phosphorus.

Question 2.
Write down construction, advantages and limitations of a solar cooker.
Answer:
Solar cooker:
Construction:
Outer surface of the solar cooker is painted with black colour and a glass plate is used as cover.
Black colour absorbs more heat and the glass plate traps solar radiation by greenhouse effect. Mirror reflects the light.

Advantages of solar cookers

• Eco friendly
• Renewable
• Used in rural areas.
• Retains all the nutrients in food due to slow cooking.

Disadvantages of solar cooker

• Silicon cells are expensive.
• Solar radiations are not uniform over the Earth’s surface.
• Cannot be used at night or on cloudy days.
• Cannot be used to make chapattis for frying as these require a temperature of 140°C or more. (Maximum temperature of 100°C only can be achieved in a solar cooker.)

Question 3.
What is hydroelectric power plant? Write down its advantages and disadvantages.
Answer:
Hydroelectric power plant: A conventional renewable source of energy is obtained from water falling from a great height. Dams are constructed to collect water flowing in high altitude rivers. The stored water has a lot of potential energy. When water is allowed to fall from a height, potential energy changes to kinetic energy, which rotates the turbines to produce electricity.

Advantages:

• It is clean and non polluting source of energy.
• Hydropower is renewable source of energy.

Disadvantages:

• Highly expensive to construct.
• Dams cannot be constructed on all river sites.
• Large areas of human habitation and agricultural fields get submerged.
  People face social and environmental problems.

Sources of Energy HOTS Questions With Answers

Question 1.
What is the role of a plane mirror and a glass sheet in a solar cooker?
Answer:
Plane mirror reflects sunlight so that maximum sunlight can enter the box.
Glass sheet traps solar radiation by greenhouse effect.

Question 2.
What kind of mirror is used in concentrating type-solar cooker?
Answer:
A concave mirror is used in concentrating type solar cooker so that maximum heat can be concentrated at a given point.

Question 3.
Name the process that produces a large amount of energy in the sun.
Answer:
Nuclear fusion

Question 4.
Name the materials used for making solar cells.
Answer:
Silicon, Germanium and Selenium

Extra Questions for Class 10 Science Chapter 14 Value Based Questions

Question 1.
A. Reddy visited his grandfather’s village Koodankular with his younger sister. They saw people protesting against building a nuclear power plant there. Reddy’s sister asked the reason for the protests. Reddy said that it could be due to the risks involved in setting up the plant near people’s habitation.
(i) Write down the advantages and disadvantages of a nuclear power plant. ‘
(ii) Values shown by A. Reddy.
(iii) Which process is used to harness nuclear energy? Explain briefly.
Answer:
(i) Advantages of Nuclear energy:

• Alternative source of energy due to depletion of fossil fuels.
• From a small amount of fuel, a large amount of energy is released.

Disadvantages of Nuclear energy:

• Hazardous nature of nuclear waste and risk of nuclear waste leakage.
• High cost of setting up a nuclear plant.
• Displacement of people from their habitat.

(ii) A. Reddy is a very intelligent person.

(iii) Nuclear energy is produced by a process called nuclear fission. During this process, the nucleus of a heavy atom (such as uranium, plutonium or thorium) when bombarded with low-energy neutrons, can be split apart into lighter nuclei. The process releases a tremendous amount of heat energy. The released energy can be used to produce steam which is used to produce electricity.

Question 2.
During summer vacations Bhaskar visited his uncle’s village. He saw a biogas plant installed by villagers for their basic energy needs.
(i) What is biogas and biogas plant?
(ii) What values are shown by villagers?
Answer:
(i) Biogas: It is a mixture of gases produced during decomposition of biomass in the absence of oxygen. Methane is the major component of biogas.
Biogas plants: Animal dung, sewage, crop residues, vegetable wastes and, poultry droppings are used to produce biogas in biogas plants.

(ii) Villagers are intelligent and care for the environment.

Question 3.
Iatin aggarwal took admission in DTO (Delhi Technological University). He saw solar devices were installed in the university.
(i) Write down a short note on solar energy and solar cooker.
(ii) Write down the values shown by the university management.
Answer:
(i) (a) Solar energy:

• Solar radiations can be converted to electricity through solar cells (photovoltaic cells).
• Photovoltaic cells convert solar radiations directly into electricity through silicon solar cells.
• Solar cells arranged on a large flat sheet to form a solar panel.

(b) Solar cooker:
Solar cookers are painted black from outside and covered with a large glass plate to trap solar radiations by greenhouse effect.

(ii) University management is concerned about energy conservation.

In this page, we are providing Human Eye and Colourful World Class 10 Extra Questions and Answers Science Chapter 11 pdf download. NCERT Extra Questions for Class 10 Science Chapter 11 Human Eye and Colourful World with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 11 Extra Questions and Answers Human Eye and Colourful World

Extra Questions for Class 10 Science Chapter 10 Human Eye and Colourful World with Answers Solutions

Extra Questions for Class 10 Science Chapter 11 Very Short Answer Type

Class 10 Science Chapter 11 Extra Questions Question 1.
Why is the colour of the clear sky blue? (NCERT Exemplar)
Answer:
Colour of the clear sky is blue: The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light.

When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour more strongly than red.

Human Eye And The Colourful World Class 10 Extra Questions And Answers Question 2.
Why do stars appear to twinkle?
Answer:
Stars appears to twinkle due to atmospheric refraction.

Human Eye Class 10 Extra Questions Question 3.
Define farthest point of an eye.
Answer:
Farthest point of an eye: The farthest point upto which the eye can see objects clearly is called far point of the eye. It is infinity for normal eye.

Human Eye And Colourful World Extra Questions Question 4.
Define power of accommodation.
Answer:
Power of accommodation: The ability of the eye lens to adjust its focal length is called accommodation.

Human Eye And The Colourful World Extra Questions Question 5.
Define least distance of distinct vision.
Answer:
Least distance of distinct vision: Minimum distance at which an object can be seen distinctly without any strain from the normal eye, i.e., 25 cm for normal vision.

Human Eye And Colourful World Class 10 Extra Questions Question 6.
Define Tyndall effect.
Answer:
Tyndall effect: The phenomenon of scattering of light by colloidal particles gives rise to Tyndall effect.
Tyndall effect can be observed when sunlight passes through a canopy of a dense forest. Here tiny droplets in mist scatters light.

Class 10 Science Ch 11 Extra Questions Question 7.
Define atmospheric refraction.
Answer:
Atmospheric refraction: If physical conditions of the refracting medium (air) are not stationary, the apparent position of the object fluctuates.

Extra Questions Of Human Eye And The Colourful World Question 8.
Why are danger signal lights red in colour?
Answer:
Danger signal lights are red in colour because red colour is least scattered by fog or smoke.

Extra Questions for Class 10 Science Chapter 11 Short Answer Type I

Cbse Class 10 Physics Human Eye And Colourful World Extra Questions Question 1.
What is meant by advance sunrise and delayed sunset? Draw a labelled diagram to explain these phenomena.
Answer:
Advance sunrise and delayed sunset is due to atmospheric refraction.
When the sun is slightly below the horizon, the sunlight coming from the less dense (vacuum) to more dense (air) medium is refracted downwards. So the sun appears to be above the horizon.

Similarly, even after actual sunset, the sun can be seen for a few minutes due to refraction of sunlight.

Class 10 Human Eye Extra Questions Question 2.
Explain formation of rainbow.
Answer:

Rainbow formation: A rainbow is a natural spectrum appearing in the sky after rain shower. It is caused by dispersion of sunlight by tiny water droplets, present in the atmosphere. The water droplets act like small prism. They refract and disperse the incident sunlight, then reflect it internally and finally refract it again.
Due to dispersion of light and internal reflection different colours appear.

Chapter 11 Science Class 10 Extra Questions Question 3.
Explain the refraction of light through a triangular glass prism using a labelled ray diagram. Hence define the angle of deviation. (NCERT Exemplar)
Answer:
Refraction of light through prism

PE – Incident ray
EF – Refracted ray
FS – Emergent ray
∠A – Angle of the prism
∠i – Angle of incidence
∠r – Angle of refraction
∠e – Angle of emergence
∠D – Angle of deviation

Refraction of light through a triangular glass prism

1. The refraction of light takes place at two surfaces firstly when light enters from air to prism and secondly when light emerges from prism.
2. Angle of prism: The angle between the two lateral faces of the prism is called angle of prism.
3. Angle of deviation: The angle between incident ray (produced forward) and emergent ray (produced backward).

Extra Questions for Class 10 Science Chapter 11 Short Answer Type II

Human Eye And The Colourful World Class 10 Questions And Answers Question 1.
What is the difference in colours of the sun observed during sunrise/sunset and noon? Give explanation for each.
Answer:
In the morning and evening, the sun lies near the horizon. Sunlight travels through a larger distance in the atmosphere and most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelength. This gives rise to the reddish appearance of the sun.
At noon sun appears white as only a little of blue and voilet colours are scattered.

Extra Questions Of Chapter 11 Science Class 10 Question 2.
Define the term dispersion of white light. Name the colour of light which bends (i) the most, (ii) the least, while passing through a glass prism. Draw a ray diagram to justify your answer.
Answer:
Dispersion of white light by a glass prism
Dispersion: The splitting of light into its component colours is called dispersion.
The red light bends the least while violet bends the most.
Spectrum: The band of the coloured components of a light beam is called spectrum.
i.e., VIBGYOR

The Human Eye And The Colourful World Extra Questions Question 3.
Explain twinkling of stars.
Answer:
Twinkling of stars:

• The twinkling of stars is due to atmospheric refraction of starlight.
• When starlight enters the earth’s atmosphere, it suffers refraction continuously. Since the physical conditions of the earth’s atmosphere are not stationary the stars appear twinkling.

Extra Questions for Class 10 Science Chapter 11 Long Answer Type

Human Eye And Colourful World Class 10 Questions With Answers Question 1.
List four common refraction defects of vision. Suggest the way of correcting these defects. (CBSE 2014)
Answer:
Defects of vision:
(i) Cataract: Crystalline lens of people at old age becomes milky and cloudy. This condition is called cataract.
It is possible to restore vision through cataract surgery.

(ii) Myopia: (Near sightedness)
A person with myopia can see nearby objects clearly but cannot see distant objects clearly.

Cause:

• Due to excessive curvature of the eye lens.
• Elongation of the eyeball.

Correction:
Concave lens of suitable power.
(a) Far point of myopic eye

(b) Myopic eye

(c) Correction for myopia

(a), (b) The myopic eye, and (c) correction for myopia with a concave lens

(iii) Hypermetropia (far-sightedness)
A person with hypermetropia can see distant objects clearly but cannot see nearby objects distinctly.

Cause:

• The focal length of the eye lens is too long.
• The eyeball has become too small.

Correction:
Convex lens of suitable power.



(a), (b) The hypermetropic eye, and (c) correction for hypermetropia

(iv) Presbyopia
The power of accommodation of the eye usually decreases with ageing. In this eye defect it is difficult to see nearby objects comfortably and distinctly without corrective eye glasses.

Cause:
Weakening of ciliary muscles and diminishing flexibility of eye lens.

Correction:
By using Bifocal lens: Upper portion consists of concave lens and lower part is convex lens.

Question 2.
Explain the structure and functioning of the human eye. How are we able to see nearby as well as distant objects?
Answer:

• Cornea: A thin membrane through which light enters the eye, maximum refraction occurs at the outer surface of cornea.
• Iris: A dark muscular membrane which controls size of pupil.
• Pupil: Regulates and controls the amount of light entering the eye.
• Eye lens: Composed of fibrous, jelly like material, with adjustable curvature, forms an inverted and real image of object at retina.
• Retina: It is a light sensitive screen on which image is formed.

The power of accommodation, that is, the ability of the eye lens to adjust its focal length, help us to see near and far objects clearly.

Human Eye and Colourful World HOTS Questions With Answers

Question 1.
How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.
Answer:
When an inverted prism is kept a little distance away from the prism causing dispersion or basically in the path of splitted beam, the spectrum recombines to form white light.

Recombination of the spectrum of white light

Question 2.
Is the position of a star as seen by us in its exact position? Justify your answer. (NCERT Exemplar)
Answer:
No, the starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position.

Question 3.
Why do we see a rainbow in the sky only after rainfall?
Answer:
The rainbow in the sky appears only after rainfall because the suspended water drops behave like prism, and refract, disperse and reflect the light rays internally.

Question 4.
A person needs a lens of power – 4.5 D for correction of his/her vision.
(a) What kind of defect in vision is he/she suffering from?
(b) What is the focal length of the corrective lens?
(c) What is the nature of the corrective lens?
Answer:
(a) Myopia
(b) f = \(\frac{1}{P}\) = \(\frac{100}{4.5}\) = 22.2 cm
(c) Concave lens

Question 5.
A narrow beam PQ of white light is passing through a glass prism ABC as shown in the diagram. (CBSE 2014)

Trace it on your answer sheet and show the path of the emergent beam as observed on the screen DE.
(i) Write the name and cause of the phenomenon observed.
(ii) Where else in nature is this phenomenon observed?
(iii) Based on this observation, state the conclusion which can be drawn about the constituents of white light.
Answer:
(i) Dispersion of light, because different colours of light bends through different angles.
(ii) Rainbow formation
(iii) White light contains seven colours i.e., VIBGYOR

Question 6.
On which factor does colour of scattered light depends?
Answer:
The colour of the scattered light depends on the size of the scattering particles. Very fine particles scatter mainly blue light while particle of larger size scatters light of longer wavelengths.

Extra Questions for Class 10 Science Chapter 11 Value Based Questions

Question 1.
Akshay, sitting in the last row in his class, could not see clearly the words written on the blackboard. When the teacher noticed it, he asked if any student sitting in the front row would volunteer to exchange his seat with Akshay. Salman immediately agreed to exchange his seat with Akshay. Akshay could now see the words written on the blackboard clearly. The teacher thought it fit to send message to Akshay’s parents advising them to get his eyesight checked.
In the context of the above event, answer the following questions:
(a) Which defect of vision is Akshay suffering from? Which type of lens is used to correct this defect?
(b) State the values displayed by the teacher and Salman.
Answer:
(a) Myopia, concave lens
(b) Teacher is very caring and knowledgeable. Salman has great concern for his friend, Akshay, and is very helpful.

Question 2.
On the rainy day, Ram reached his grandfather’s place in village. On the way to the house he saw a beautiful rainbow in the sky. In the night, he saw lots of twinkling stars in the clear sky. He was very excited to see these beautiful natural phenomenon, which he was not able to see in the city, where he lived with his father. Explain the phenomenon on the basis of science. Do you think that pollution in the atmosphere affects the formation of rainbow and twinkling of stars. Do you agree with the fact that pollution free environment will strengthen such natural phenomenon in the cities as well. Elaborate.
Answer:

• The twinkling of star is due to atmospheric refraction.
• The formation of rainbow is due to dispersion, refraction and internal reflection.
• Yes, pollution in atmosphere affects the formation of rainbow and twinkling of stars.

Question 3.
Vinay’s father cannot read a book placed 25 cm from his eye. But he feels a little comfortable when the book is placed 50 cm away. Vinay adviced his father for checkup of the eyes.
(a) From which defect of vision Vinay’s father may be suffering from? Give the proper correction.
(b) State the values of Vinay.
Answer:
(a) Hypermetropia, correction can be done by using suitable convex lens.
(b) Vinay is caring and intelligent.

Question 4.
Ankit’s grandma is facing a problem of clouded, blurred and dim vision. Ankit took her to the doctor.
(a) From which defect of vision Ankit’s grandma may be suffering from? Give proper correction.
(b) State the values of Ankit.
Answer:
(a) Cataract. It is possible to restore vision through cataract surgery.
(b) Ankit is caring and intelligent.

Here we are providing Pair of Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers

Quadratic Equations Class 10 Extra Questions Very Short Answer Type

Quadratic Equation Class 10 Extra Questions Question 1.
What will be the nature of roots of quadratic equation 2x² + 4x – n = 0?
Solution:
D = b² – 4ac
⇒ 4² – 4 x 2 (-7)
⇒ 16 + 56 = 72 > 0
Hence, roots of quadratic equation are real and unequal.

Quadratic Equation Extra Questions Question 2.
If \(\frac{1}{2}\) is a root of the equation x² + kx – \(\frac{5}{4}\) = 0, then find the value of k.
Solution:
∴ \(\frac{1}{2}\) is a root of quadratic equation.
∴ It must satisfy the quadratic equation.

Quadratic Equations Class 10 Extra Questions Question 3.
If ax² + bx + c = 0 has equal roots, find the value of c.
Solution:
For equal roots D = 0
i.e., b² – 4ac = 0
⇒ b² = 4 ac
⇒ c = \(\frac{b^{2}}{4 a}\)

Quadratic Equation Class 10 Questions Question 4.
If a and b are the roots of the equation x² + ax – b = 0, then find a and b.
Solution:
Sum of the roots = a + b = – \(\frac{B}{A}\) = – a
Product of the roots = ab = \(\frac{B}{A}\) = – b
= a + b = – a and ab = -b
⇒ 2a = -b and a = -1
⇒ b = 2 and a = -1

Class 10 Maths Chapter 4 Extra Questions With Solutions Question 5.
Show that x = – 2 is a solution of 3x² + 13x + 14 = 0.
Solution:
Put the value of x in the quadratic equation,
⇒ LHS = 3x² + 13x + 14
⇒ 3(-2)² + 13(-2) + 14
⇒ 12 – 26 + 14 = 0
⇒ RHS Hence, x = -2 is a solution.

Quadratic Equation Class 10 Extra Questions With Answers Question 6.
Find the discriminant of the quadratic equation 4√2x² + 8x + 2√2 = 0).
Solution:
D = 62 – 4ac = (8)² – 4(4√2)(2√2)
⇒ 64 – 64 = 0

Quadratic Equations Class 10 Extra Questions Short Answer Type 1

Class 10 Quadratic Equation Extra Questions Question 1.
State whether the equation (x + 1)(x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution:
(x + 1)(x – 2) + x = 0
⇒ x² – x – 2 + x = 0
⇒ x² – 2 = 0
D = b² – 4ac
⇒ (-4(1)(-2) = 8 > 0
∴ Given equation has two distinct real roots.

Extra Questions Of Quadratic Equations Class 10 Question 2.
Is 0.3 a root of the equation x² – 0.9 = 0? Justify.
Solution:
∵ 0.3 is a root of the equation x² – 0.9 = 0
∴ x² – 0.9 = (0.3)² – 0.9 = 0.09 – 0.9 ≠ 0
Hence, 0.3 is not a root of given equation.

Quadratic Equations Extra Questions Question 3.
For what value of k, is 3 a root of the equation 2x² + x + k = 0?
Solution:
3 is a root of 2x² + x + k = 0, when
⇒ 2(3)² + 3 + k = 0
⇒ 18+ 3 + k = 0
⇒ k = – 21

Quadratic Equation Class 10 Important Questions With Solutions Question 4.
Find the values of k for which the quadratic equation 9x² – 3kx + k = 0 has equal roots.
Solution:
For equal roots:
D = 0
⇒ b² – 4ac = 0
⇒ (- 3k)² – 4 × 9 × k = 0
⇒ 9k² = 36k
⇒ k = 4

Class 10 Quadratic Equations Extra Questions Question 5.
Find the value of k for which the equation x² + k(2x + k – 1)+ 2 = 0 has real and equal roots.
Solution:
Given quadratic equation: x² + k(2x + k-1) + 2 = 0)
= x² + 2kx + (k² – k + 2) = 0
For equal roots, b² – 4ac = 0
⇒ 4k² – 4k² + 4k – 8 = 0
⇒ 4k = 8
⇒ k = 2

Quadratic Equation Questions For Class 10 Question 6.
If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then find the value of k.
Solution:
Since – 5 is a root of the equation 2x² + px – 15 = 0
∴ 2(-5)² + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p(x² + x) + k = 0 or 7x² + 7x + k = 0 has equal roots
∴ D = 0
i.e., b² – 4ac = 0 or 49- 4 × 7k = 0
⇒ k = \(\frac{49}{28}\) = \(\frac{7}{4}\)

Extra Questions On Quadratic Equations Class 10 Question 7.
Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Solution:
Yes, x² – 4x + 1 = 0 is a quadratic equation with rational co-efficients.

Questions On Quadratic Equations Class 10 Question 8.
Write the set of values of k for which the quadratic equation 2x² + kx + 8 = 0 has real roots.
Solution:
For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ k² – 4(2)(8) ≥ 0
⇒ k² – 64 ≥ 0
⇒ k² ≥ 64
⇒ k ≤ -8 and k ≥ 8

Class 10 Maths Quadratic Equations Extra Questions Question 9.
Solve the quadratic equation 2x² + ax – a² = 0 for x.
Solution:
2x² + ax – a² = 0
Here, a = 2, b = a and c = -a².
Using the formula,

Class 10 Quadratic Equations Important Questions Question 10.
Find the roots of the quadratic equation
Solution:
The given quadratic equation is
√2x² + 7x + 5√2 = 0
By applying mid term splitting, we get
√2x² + 2x + 5x + 5√2 = 0
⇒ √2x(x + √2) + 5(x + √2)
⇒ (√2x + 5) + 5(x + √2) = 0
⇒ x = \(\frac{-5}{\sqrt{2}}/latex], -√2 or [latex]\frac{-5 \sqrt{2}}{2}\), -√2

Quadratic Equation Class 10 Extra Questions Pdf Question 11.
Find the values of p for which the quadratic equation 4x² + px + 3 = 0 has equal roots.
Solution:
For equal roots;
D = 0
⇒ b² – 4ac = 0
⇒ p² – 4 × 4 × 3 = 0
⇒ p² – 48 = 0
⇒ p² = 48
⇒ p = ± √48
⇒ p = 4√3 or -4√3

Ch 4 Maths Class 10 Extra Questions Question 12.
Solve for x: √13x? – 2√3x – 2√3 = 0
Solution:
√3x² – 2√3x – 2√3 = 0
⇒ √3x² – 3√2x + √2x – 2√3 = 0
⇒ √3x(x – √6) + √2(x – √6) = 0
⇒ (√3x + 2)(x – √6) = 0
⇒ √3x + √2 = 0 or x – √6 = 0
⇒ x = \(\frac{-√2}{√3}\) or x = √6

Chapter 4 Maths Class 10 Extra Questions Question 13.
If x = \(\frac{2}{3}\) and x = -3 are roots of the quadratic equation ax² + 7x + b = 0, find the values of a and b.
Solution:
Let us assume the quadratic equation be Ax² + Bx + C = 0.
Sum of the roots = –\(\frac{B}{A}\)

Class 10 Maths Ch 4 Extra Questions Question 14.
A two-digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
Solution:
Let the ten’s digit be x and unit’s digit = y
Number 10x + y

Extra Questions For Quadratic Equation Class 10 Question 15.
Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.
Solution:
Let the roots of the given equation be a and 6α.
Thus the quadratic equation is (x – a) (x – 6α) = 0
⇒ x² – 7αx + 6α² = 0 …(i)
Given equation can be written as

Question 16.
If ad ≠ bc, then prove that the equation
(a² + b²) x² + 2(ac + bd)x + (c²+ d²) = 0 has no real roots.
Solution:
The given quadratic equation is (a² + b²⁾x² + 2(ac + bd)x +(c²+ d²) = 0
D = b² – 4ac
= 4(ac + bd)² – 4(a² + b²) (c²+ d²)
= -4(a²d² + b²c²– 2abcd) = – 4(ad – bc)²
Since ad ≠ bc
Therefore D < 0
Hence, the equation has no real roots.

Question 17.
Solve for x: √13x² – 2x – 8√3 = 0
Solution:
√3x² – 2x – 8√3 = 0
By mid term splitting
⇒ √3x² – 6x + 4x – 8√3 = 0
⇒ √3x(x – 2/3) + 4 (x – 2/3) = 0
⇒ (x – 2√3)(√3x + 4) = 0
⇒ Either (x – 2√3) = 0 or (√3x + 4) = 0
⇒ x = \(\frac{-4}{\sqrt{3}}\), 2√3

Quadratic Equations Class 10 Extra Questions Short Answer Type 2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) √2x² + 7x + 5√2 = 0 (ii) 2x² – x + \(\frac{1}{8}\) = 0
Solution:
(i) We have, √2x² + 7x + 5√2 = 0
= √2x² + 5x + 2x + 5√2 = 0
x(√2x + 5) + √2 (√2x + 5) = 0
= (√2x + 5)(x + √2) = 0
∴ Either √2x + 5 = 0 or x + √2 = 0
∴ x = – \(\frac{5}{\sqrt{2}}\) or x = -√2
Hence, the roots are – \(\frac{5}{\sqrt{2}}\) and -√2.
(ii) We have, 2x² – x + \(\frac{1}{8}\) = 0

Question 2.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² + x – 4 = 0
(ii) 4x² + 4√3x + 3 = 0
Solution:
(i) We have, 2x² + x – 4 = 0
On dividing both sides by 2, we have

(ii) We have, 4x² + 4√3x + 3 = 0

Question 3.
Find the roots of the following quadratic equations by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0
(ii) 4x² + 4√3x + 3 = 0
Solution:
(i) We have, 2x² – 7x + 3 = 0
Here, a = 2, b = -7 and c = 3
Therefore, D = b² – 4ac
⇒ D = (-7)² – 4 × 2 × 3 = 49 – 24 = 25
∵ D > 0, ∴ roots exist.


So, the roots of given equation are 3 and \(\frac{1}{2}\)

(ii) We have, 4x² + 4√3x + 3 = 0
Here, a = 4, b = 4√3 and c = 3
Therefore, D = b² – 4ac = (4√3)² – 4 × 4 × 3 = 48 – 48 = 0
∴ D = 0, roots exist and are equal.

Question 4.
Using quadratic formula solve the following quadratic equation:
p²x² + (p² – q²) x – q² = 0
Solution:
We have, p²x² + (p² – q²) x – q² = 0
Comparing this equation with ax² + bx + c = 0, we have
a = p², b = p² – q² and c = – q²
∴ D = b² – 4ac
⇒ (p² – q)² – 4 × p² × (-q²)
⇒ (p² – q²)² + 4p²q²
⇒ (p² + q³)² > 0
So, the given equation has real roots given by

Question 5.
Find the roots of the following equation:

Solution:

⇒ (x + 3) (x – 6)
⇒ -20 or x² – 3x + 2 = 0
⇒ x² – 2x -x + 2 = 0
⇒ x(x – 2) -1(x – 2) = 0)
⇒ (x – 1) (x – 2) = 0
⇒ x = 1 or x = 2
Both x = 1 and x = 2 are satisfying the given equation. Hence, x = 1, 2 are the solutions of the equation.

Question 6.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 3x² – 4√3x + 4 = 0) (ii) 2x² – 6x + 3 = 0
Solution:
(i) We have, 3x² – 4√3x + 4 = 1
Here, a = 3, b = – 4√3 and c = 4
Therefore,
D = b² – 4ac
⇒ (- 4√3)² – 4 × 3 × 4
⇒ 48 – 48 = 0
Hence, the given quadratic equation has real and equal roots.

(ii) Wehave, 2x² – 6x + 3 = 0
Here, a = 2, b = -6, c = 3
Therefore, D = b² – 4ac
= (-6)² 4 × 2 × 3 = 36 – 24 = 12 > 0
Hence, given quadratic equation has real and distinct roots.

Question 7.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Solution:
(i) We have, 2x² + kx + 3 = 0
Here, a = 2, b = k, c = 3
D = b² – 4ac = k² – 4 × 2 × 3 = k² – 24 For equal roots
D = 0
i.e., k² – 24 = 0
⇒ ķ² = 24
⇒ k = ± √24
⇒ k = + 2√6

(ii) We have, kx(x – 2) + 6 = 0
⇒ kx² – 2kx + 6 = 0
Here, a = k, b = – 2k, c = 6
For equal roots, we have
D = 0
i.e., b² – 4ac = 0
⇒ (-2k)² – 4 × k × 6 = 0
⇒ 4k² – 24k = 0
⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k = 0 6)ecause if k = 0 then given equation will not be a quadratic equation).
So, k = 6.

Question 8.
If the roots of the quadratic equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.
Solution:
Since the equation (a – b)x² + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)² – 4(a – b) (c – a) = 0
⇒ b² + c² – 2bc – 4(ac – a² – bc + ab)
⇒ b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0
⇒ 4a² + b² + c² – 4ab + 2bc – 4ac = 0
⇒ (2a)² + (- b)² + (-c)² + 2(2a) (-b) + 2(-b) (-c) + 2(-c) 2a = 0
⇒ (2a – b – c)² = 0
⇒ 2a – b – c = 0
⇒ 2a = b + c. Hence Proved

Question 9.
If the equation (1 + m²)x² + 2mcx + c²– a² = 0 has equal roots, show that c² = a² (1 + m²).
Solution:
The given equation is (1 + m²) x² + (2mc) x + (c²– a²) = 0
Here, A = 1 + m², B = 2mc and C = c² – a²
Since the given equation has equal roots, therefore D = 0 = B² – 4AC = 0.
⇒ (2mc)² – 4(1 + m²) (c² – a²) = 0
⇒ 4m²c² – 4(c² – a² + m²c² – m²a²) = 0
⇒ m²c² – c² + a² – m²c² + m²a² = 0. [Dividing throughout by 4]
⇒ – c² + a² (1 + m²) = 0
⇒ c² = a(1 + m²) Hence Proved

Question 10.
If sin θ and cos θ are roots of the equation ax² + bx + c = 0, prove that a² – b² + 2ac = 0.
Solution:

Question 11.
Determine the condition for one root of the quadratic equation ax² + bx + c = 0 to be thrice the other.
Solution:
Let the roots of the given equation be a and 3α.

Question 12.
Salve for

Solution:

⇒ (4x – 2)(2x – 1) – (3x + 9)(x + 3) = 5(x + 3)(2x – 1)
⇒ (8x² – 4x – 4x + 2) – (3x² + 9x + 9x + 27) = 5(2x² – x + 6x – 3)
⇒ 8x² – 8x + 2 – 3x² – 18x – 27 = 10x² + 25x – 15
⇒ 5x² – 26x – 25 = 10x² + 25x – 15
⇒ 5x² + 51x + 10 = 0
⇒ 5x² + 50x + x + 10 = 0
⇒ 5x (x + 10) + 1 (x + 10) = 0
⇒ (5x + 1) (x + 10) = 0
⇒ 5x + 1 = 0 or x + 10 = 0
⇒ x = \(\frac{-1}{5}\) or x = -10

Question 13.
Solve the equation

Solution:

⇒ (4 – 3x) (2x + 3) = 5x ⇒ 8x – 6x² + 12 – 9x = 5x
⇒ 6x² + 6x – 12 = 0
⇒ x² + x – 2 = 0
⇒ x² + 2x + x – 2 = 0
⇒ x(x + 2)-1(x + 2) = 0
⇒ (x – 1)(x + 2) = 0
⇒ x – 1 = 0 or x + 2 = 0
⇒ x = 1 or x = -2

Question 14.
Solve for

Solution:

⇒ (16 – x) (x + 1) = 15x
⇒ 16x – x² + 16 – x = 15x
⇒ x² + 15x – 15x – 16 = 0
⇒ x² = 16
⇒ x = ± 4

Question 15.
Solve for x: + 5x – (a² + a – 6) = 0
Solution:
⇒ x² + 5x – (a² + a – 6) = 0
⇒ x² + 5x – (a? + 3a – 2a – 6) = 0
⇒ x² + 5x – [a(a + 3) -2 (a + 3)] = 0
⇒ x² + 5x – (a – 2) (a + 3) = 0
∴ x² + (a + 3)x – (a – 2) x – (a – 2) (a + 3) = 0
⇒ x[x + (a + 3)]-(a – 2) [X + (a + 3)] = 0
⇒ [{x + (a + 3)} {x – (a – 2)}] = 0
∴ x = -(a + 3) or x = (a -2)
⇒ -(a + 3), (a – 2)
Alternative method
x² + 5x -(a² + a – 6) = 0

Question 16.
Solve for

Solution:

⇒ 2x(2x + 3) + (x – 3) + (3x + 9) = 0
⇒ 4x² + 10x + 6 = 0
⇒ 2x² + 5x + 3 = 0
⇒ (x + 1) (2x + 3) = 0
⇒ x = -1, x = – \(\frac{3}{2}\)
But x ≠ – \(\frac{3}{2}\)
∴ x = -1

Question 17.
Solve for

Solution:

⇒ 3(x – 3 + x – 1) = 2(x – 1)(x – 2)(x – 3)
⇒ 3(2x – 4) = 2(x – 1)(x – 2)(x – 3)
⇒ 3 × 2(x – 2) = 2(x – 1)(x − 2)(x – 3)
⇒ 3 = (x – 1)(x – 3) i.e., x² – 4x = 0
⇒ x(x – 4) = 0 ∴ x = 0, x = 4

Question 18.
If the roots of the equation (c- ab)x² – 2(a²– bc)x + b² – ac = 0 in x are equal, then show that either a = 0 or a³ + b³ + c³ = 3abc.
Solution:
For equal roots D = 0
Therefore 4(a² – bc)² – 4(c² – ab) (b² – ac) = 0
⇒ 4[a⁴ + b²c² – 2a²bc – b²c² + ac³ + ab³ – a²bc] = 0.
⇒ a(a³ + b³ + c³ – 3abc) = 0
Either a = 0 or a³ + b³ + c³ = 3abc

Question 19.
If the roots of the quadratic equation
(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
are equal, then show that a = b = c.
Solution:
Given (x – a) (x – b) + (x – b) (x – c) + (x – 6) (x – a) = 0
⇒ x² – ax – bx + ab + x² – bx – cx + bc + x² – cx – ax + ac = 0
⇒ 3x² – 2(a + b + c)x + ab + bc + ca = 0
Now, for equal roots D = 0
⇒ B² – 4AC = 0
⇒ 4(a + b + c)² – 12(ab + bc + ca) = 0
4a² + 4b² + 4c² + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0
⇒ 2[2a² + 2b² + 2co – 2ab – 2bc – 2ca] = 0
⇒ 2[(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca)] = 0
⇒ [(a – b)² + (b – c)² + (c – a)²] = 0
⇒ a – b = 0, b – c = 0, c – a = 0
⇒ a = b, b = c, c = a
⇒ a = b = c

Quadratic Equations Class 10 Extra Questions Long Answers

Question 1.
Using quadratic formula, solve the following equation for x:
abx² + (b² – ac) x – bc = 0
Solution:
We have, abx² + (b² – ac) x – bc = 0
Here, A = ab, B = b² – ac, C = – bc

Question 2.
Find the value of p for which the quadratic equation
(2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.
Solution:
Since the quadratic equation has equal roots, D = 0
i.e., b² – 4ac = 0
In (2p + 1 )x² – (7p + 2)x + (7p – 3) = 0
Here, a = (2p + 1), b = -(7p + 2), c = (7p – 3)

Question 3.
Solve for

Solution:

Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is Find his present age.
Solution:
Let the present age of Rehman be x years.
So, 3 years ago, Rehman’s age = (x – 3) years
And 5 years from now, Rehman’s age = (x + 5) years
Now, according to question, we have

But x ≠ -3 (age cannot be negative)
Therefore, present age of Rehman = 7 years.

Question 5.
The difference of two natural numbers is 5 and the difference of their reciprocals is \(\frac{1}{10}\). Find the numbers.
Solution:
Let the two natural numbers be x and y such that x > y.
According to the question,
Difference of numbers, x – y = 5 ⇒ x = 5 + y …..(i)
Difference of the reciprocals,

∴ y is a natural number.
∵ y = 5
Putting the value of y in (i), we have
⇒ x = 5 + 5
⇒ x = 10
The required numbers are 10 and 5.

Question 6.
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Solution:
Let the two consecutive odd numbers be x and x + 2.

Hence, the numbers are 13 and 15 or -15 and -13.

Question 7.
The sum of two numbers is 15 and the sum of their reciprocals is 3. Find the numbers.
Solution:
Let the numbers be x and 15 – x.
According to given condition,

⇒ 150 = 3x(15 – x)
⇒ 50 = 15x – x²
⇒ x² – 15x + 50 = 0
⇒ x² – 5x – 10x + 50 = 0
⇒ x(x – 5) -10(x – 5) = 0
⇒ (x – 5)(x – 10) = 0
⇒ x = 5 or 10.
When x = 5, then 15 – x = 15 – 5 = 10
When x = 10, then 15 – x = 15 – 10 = 5
Hence, the two numbers are 5 and 10.

Question 8.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics be x.
Therefore, Shefali’s marks in English is (30 – x).
Now, according to question,
⇒ (x + 2) (30 – x – 3) = 210
⇒ (x + 2) (27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ 25x – x² + 54 – 210 = 0
⇒ 25x – x² – 156 = 0
⇒ -(x² – 25x + 156) = 0
⇒ x² – 25x + 156 = 0
= x² – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13) (x – 12) = 0
Either x – 13 or x – 12 = 0
∴ x = 13 or x = 12
Therefore, Shefali’s marks in Mathematics = 13
Marks in English = 30 – 13 = 17
or Shefali’s marks in Mathematics = 12
marks in English = 30 – 12 = 18.

Question 9.
A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let the uniform speed of the train be x km/h.

But x cannot be negative, so x ≠ – 45
therefore, x = 40
Hence, the uniform speed of train is 40 km/h.

Question 10.
The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let x be the length of the side of first square and y be the length of side of the second square.
Then, x² + y² = 468 …(i)
Let x be the length of the side of the bigger square.
4x – 4y = 24
⇒ x – y = 6 or x = y + 6 …(ii)
Putting the value of x in terms of y from equation (ii), in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468 or 232 + 12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
Either y + 18 = 0 or y – 12 = 0
⇒ y = -18 or y = 12
But, sides cannot be negative, so y = 12
Therefore, x = 12 + 6 = 18
Hence, sides of two squares are 18 m and 12 m.

Question 11.
Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence, Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Solution:
Seven years ago, let Swati’s age be x years. Then, seven years ago Varun’s age was 5x² years.
∴ Swati’s present age = (x + 7) years
Varun’s present age = (5x² + 7) years
Three years hence,
Swati’s age = (x + 7 + 3) years = (x + 10) years
Varun’s age (5x² + 7 + 3) years = (5x² + 10) years
According to the question,

Hence, Swati’s present age = (2 + 7) years = 9 years
and Varun’s present age = (5 × 2² + 7) years = 27 years

Question 12.
A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.
Solution:
Let the original speed of the train = x km/h.


Therefore, the usual speed of the train = 25 km/h.

Question 13.
A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Let the digit at tens place be x.

Question 14.
If twice the area of a smaller square is subtracted from the area of a larger square; the result is 14 cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of the two squares.
Solution:
Let the sides of the larger and smaller squares be x and y respectively. Then
x² – 2y² = 14 …..(i)
and 2x² + 3y² = 203 ……(ii)
Operating (ii) -2 × (i), we get
⇒ 2x² + 3y² – (2x² – 4y²) = 203 – 2 × 14
⇒ 2x² + 3y² – 2x² + 4y² = 203 – 28
⇒ 7y² = 175
⇒ y² = 25
⇒ y ± 15
⇒ y = 5 [∵ Side cannot be negative]
By putting the value of y in equation (i), we get
x² – 2 × 5 = 14
⇒ x² – 50 = 14 or x² = 64
∴ x = 8 or x = 8
∴ Sides of the two squares are 8 cm and 5 cm.

Question 15.
If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Solution:
Let the present age of Zeba be x years
Age before 5 years = (x – 5) years According to given condition,
⇒ (x – 5)² = 5x + 11
⇒ x² + 25 – 10x = 5x + 11
⇒ x² – 10x – 5x + 25 – 11 = 0
⇒ x² – 15x + 14 = 0
⇒ x² – 14x – x + 14 = 0
⇒ x(x – 14) -1(x – 14) = 0
⇒ (x – 1)(x – 14) = 0
⇒ x – 1 = 0 or x – 14 = 0
x = 1 or x = 14
But present age cannot be 1 year.
∴ Present age of Zeba is 14 years.

Question 16.
A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/h.
Speed of motorboat in still water = 18 km/h
Speed of motorboat in upstream = (18 – x) km/h
Speed of motorboat in downstream = (18 + x) km/h
Distance travelled = 24 km.

Speed of motorboat = 6 km/h. (∵Speed cannot be negative)

Question 17.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Solution:
Let the natural number be x
According to the question,
⇒  x + 12 = \(\frac{60}{x}\)
= x² + 12x – 160 = 0
= x² + 20x – 8x – 160 = 0
= x(x + 20) -8(x + 20) = 0
= (x + 20) (x – 8) = 0
x = -20 (Not possible) or x = 8
Hence, the required natural number is 8.

Question 18.
The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Solution:
Let the two consecutive multiples of 7 be x and x +7.
x² + (x + 7)² = 637
= x² + x² + 49 + 14x = 637
= 2x² + 14x – 588 = 0
= x² + 7x – 294 = 0
= x² + 21x – 14x – 294 = 0
= x(x + 21) – 14(x + 21) = 0
= x(x + 14) (x + 21) = 0
= x = 14 or x = -21
The multiples are 14 and 21.

Question 19.
Solve for

Solution:

Question 20.
Find the positive value(s) of k for which both quadratic equations x² + kx + 64 = 0 and x² – 8x + k = 0 will have real roots.
Solution:
(i) For x² + kx + 64 = 0 to have real roots
⇒ k² – 4(1)(64) ≥ 0 i.e., k² – 256 ≥ 0
⇒ k ≥ ± 16

(ii) For x² – 8x + k = 0 to have real roots
⇒ (-8)² – 4(k) ≥ 0 i.e., 64 – 4k ≥ 0
⇒ k ≤ ± 16
For (i) and (ii) to hold simultaneously k = 16

Question 21.
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Solution:
Let the speed of stream be x km/h.
∴ Speed of boat upstream = (15 – x) km/h.
Speed of boat downstream = (15 + x) km/h.
According to question,

⇒ 30 × 2 × 30 = 9(225 – x²)
⇒ 100 × 2 = 225 – x²
⇒ 200 = 225 – x²
⇒ x² = 25
⇒ x = ±5
⇒ x = 5 (Rejecting – 5)
∴ Speed of stream = 5 km/h

Question 22.
Ram takes 6 days less than Bhagat to finish a piece of work. If both of them together can finish the work in 4 days, in how many days Bhagat alone can finish the work.
Solution:
Let Bhagat alone can do the work in x number of days
∴ Ram takes (x – 6) number of days
Work done by Bhagat in 1 day = \(\frac{1}{x}\)
Work done by Ram in 1 day = \(\frac{1}{x-6}\)
According to the question,

Quadratic Equations Class 10 Extra Questions HOTS

Question 1.
One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Solution:
Let x be the total number of camels.
Then, number of camels in the forest = \(\frac{x}{4}\)
Number of camels on mountains = 2√x
and number of camels on the bank of river = 15

But, the number of camels cannot be a fraction.
∴ y = 6
⇒ x = x² = 36
Hence, the number of camels = 36

Question 2.
Solve the following quadratic equation:
9x² – 9 (a + b) x + [2a² + 5ab + 2b²] = 0.
Solution:
Consider the equation 9x² – 9 (a + b) x + [2a² + 5ab + 2b²] = 0)
Now comparing with Ax² + Bx + C = 0, we get
A = 9, B = -9 (a + b) and C = [2a² + 5ab + 2b²]
Now discriminant, .
D = B² – 4AC

Question 3.
Two taps running together can fill a tank in 3 hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?
Solution:
Let, time taken by faster tap to fill the tank be x hours
Therefore, time taken by slower tap to fill the tank = (x + 3) hours
Since the faster tap takes x hours to fill the tank.

Hence, time taken by faster tap to fill the tank = x = 5 hours
and time taken by slower tap = x + 3 = 5 + 3 = 8 hours.

Question 4.
In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed so that the area of grass strip of uniform width surrounding the pond would be 1184 m². Find the length and breadth of the pond.
Solution:

Let ABCD be rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass area, which is same around the pond.
Given, Length of lawn = 50 m
Width of lawn = 40 m
Length of pond = (50 – 2x)m
Breadth of pond = (40 – 2x)m
Also given,
Area of grass surrounding the pond = 1184 m²
⇒ Area of rectangular lawn – Area of pond = 1184 m²
⇒ 50 × 40 – {(50 – 2x) × (40 – 2x)} = 1184
⇒ 2000 – (2000 – 80x – 100x + 4x²) = 1184
⇒ 2000 – 2000 + 180x – 4x² = 1184
⇒ 4x² – 180x + 1184 = 0
⇒ x² – 45x + 296 = 0
⇒ x² – 37x – 8x + 296 = 0
⇒ x(x – 37) – 8(x – 37) = 0
⇒ (x – 37) (x – 8) = 0
⇒ x- 37 = 0 or x – 8 = 0)
⇒ x = 37 or x = 8
x = 37 is not possible as in this case length of pond becomes 50 – 2 × 37 = -24 (not possible) Hence, x = 8 is acceptable
∴ Length of pond = 50 – 2 × 8 = 34 m
Breadth of pond = 40 – 2 × 8 = 24 m

Question 5.
A car covers a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.
Solution:
Let speed of the car be x km/h
According to question
Time taken = \(\frac{x}{2}\) h.

⇒ x = 72 km/h [Taking square root both sides]
∴ Time taken = \(\frac{x}{2}\) = 36 hours.

In this page, we are providing Periodic Classification of Elements Class 10 Extra Questions and Answers Science Chapter 5 pdf download. NCERT Extra Questions for Class 10 Science Chapter 5 Periodic Classification of Elements with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 5 Extra Questions and Answers Periodic Classification of Elements

Extra Questions for Class 10 Science Chapter 5 Periodic Classification of Elements with Answers Solutions

Extra Questions for Class 10 Science Chapter 5 Very Short Answer Type

Periodic Classification Of Elements Class 10 Extra Questions With Answers Question 1.
Name two elements whose properties were predicted on the basis of their positions in Mendeleev’s periodic table.
Answer:
Gallium (Eka-aluminium) and germanium (Eka-silicon).

Periodic Classification Of Elements Extra Questions Question 2.
Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeleev.  [NCERT Exemplar]
Answer:
Eka-aluminium represents gallium (Ga) with valency three and Eka-silicon is for germanium (Ge) with valency four. The formulae of their respective chlorides are GaCl₃ and GeCl₄.

Class 10 Science Chapter 5 Extra Questions Question 3.
How many elements are there in period 2?
Answer:
Eight (8).

Periodic Classification Of Elements Class 10 Extra Questions Question 4.
Name all the elements present in group-17 of the Modern Periodic Table.
Answer:
F, Cl, Br, I, At.

Periodic Classification Of Elements Class 10 Questions And Answers Question 5.
Name two elements whose atomic weight were corrected on the basis of their position in Mendeleev’s periodic table.
Answer:
Gold (Au) and platinum (Pt).

Extra Questions Of Periodic Classification Of Elements Question 6.
Arrange the following elements in the increasing order of their metallic character.
Mg, Ca, K, Ge, Ga     [NCERT Exemplar]
Answer:
Ge < Ga Mg < Ca < K.

Class 10 Periodic Classification Of Elements Extra Questions Question 7.
Name the scientist who proposed the Modern Periodic Law.
Answer:
Henry Moseley, a scientist, proposed the Modern Periodic Law.

Chapter 5 Science Class 10 Extra Questions Question 8.
What is meant by a group in the periodic table?
Answer:
Groups are the vertical columns/vertical lines in the periodic table.

Ch 5 Science Class 10 Extra Questions Question 9.
What is meant by a period in the periodic table?
Answer:
Periods are the horizontal rows in the periodic table.

Class 10 Science Ch 5 Extra Questions Question 10.
Name the elements present in period 1 of the modern periodic table.
Answer:
Hydrogen (H) and Helium (He).

Extra Questions Of Chapter 5 Science Class 10 Question 11.
An element A is in group II (group 2) of the periodic table:
(а) What will be the formula of its chloride?
(b) What will be the formula of its oxide?
Answer:
(a) ACl₂
(b) AO.

Extra Questions Of Periodic Classification Of Elements Class 10 Question 12.
Why are the elements calcium, strontium and barium named as alkaline earths?
Answer:
These elements are called alkaline earths because their oxides are alkaline in nature and exist in the earth.

Extra Question Of Periodic Classification Of Elements Question 13.
Give the name and electronic configuration of second alkali metal.
Answer:
The second alkali metal is sodium (Na). Its electronic configuration is 2, 8, 1.

Class 10 Chapter 5 Science Extra Questions Question 14.
What are Metalloids?
Answer:
Those elements which show some properties of both metals and non-metals are called metalloids. Examples: Boron, Silicon, Germanium, Arsenic, Antimony, Tellurium and Polonium.

Periodic Classification Extra Questions Question 15.
Name three elements whose atomic masses were correct on the basis of their positions in Mendeleev’s periodic table.
Answer:

1. Beryllium (Be, group II A)
2. gold (Au, group II B)
3. platinum (Pt, group VIII).

Question 16.
Name three elements which behave as metalloids.
Answer:
The elements are:

1. arsenic (As)
2. antimony (Sb)
3. germanium (Ge).

Question 17.
Name the inert gas which has two electrons in its valence shell.
Answer:
Helium.

Question 18.
Name the most metallic and most non-metallic element in the Periodic Table.
Answer:
The most metallic elements is francium (group 1) and most non-metallic element is fluorine (group 17).

Question 19.
In the Modern Periodic Table, which are the metals among the first ten elements?
Answer:
Lithium (Li), Beryllium (Be) and Boron (B) are the metal elements among first ten elements in the Modern Periodic Table.

Question 20.
Two elements A and B belong to the same period. What is common in them?
Answer:
They have the same number of shells.

Question 21.
The electronic configuration of an element is 2, 8, 6. Identify the element and name of the family to which it belongs.
Answer:
The element with configuration 2, 8, 6 (Z = 16) is sulphur. It belongs to the oxygen family.

Question 22.
Calcium, Strontium and Barium form a Dobereiner’s triad. The atomic masses of calcium and Barium are 40 and 137 respectively. Predict the atomic mass of strontium.
Answer:
According to Dobereiner’s triad, atomic mass of strontium is the arithmetic means of Ca and Ba as strontium lies between Ca and Ba.
Atomic mass of Sr = \(\frac{40+137}{2}=\frac{177}{2}=88.5\)

Question 23.
The formula of magnesium oxides is MgO. Write the formula of magnesium chloride.
Answer:
Oxygen is divalent in nature. The valency of magnesium in magnesium oxide is + 2. The formula of magnesium chloride is MgCl₂ since chlorine has valency equal to one.

Question 24.
State whether the following statement is true or false?
The valency of an element of group 17 is 7.
Answer:
This statement is wrong. The number of valence electrons in an element of group 17 = 17 – 10 = 7.
Therefore, the valency of the element = 8 – 7 = 1.

Question 25.
Name two elements whose atomic weight were corrected on the basis of their position in Mendeleev periodic table.
Answer:

1. Gold (Au)
2. Platinum (Pt).

Question 26.
The two isotopes of chlorine have atomic mass 35.4 and 37.4. Should they be placed in separate slots in the modern periodic table?
Answer:
No, they should be placed in the same slot because the modern periodic table is based on the atomic number of the elements.

Question 27.
List any two properties of the elements belonging to the first group of the modem periodic table.   [Delhi 2014]
Answer:
Elements of first group have one valence electron each in their atoms. All the elements of group 1 have the same valency of 1.

Question 28.
What is the basic difference in the electronic configuration of the elements belonging to group 1 and group 2?
Answer:
All elements belonging to group 1 have one electron in the valence shell while all elements belonging to group 2 have two electrons in their valence shell.

Question 29.
Element ‘Y’ with atomic number 3 combines with element ‘A’ with atomic number 17. What will be the formula of the compound?
Answer:
The electronic distribution in elements ‘A’ and ‘Y’ are 2, 1 and 2, 8, 7 respectively. Both have valency equal to 1. The formula of the compound is AY.

Question 30.
Name the property which remains unchanged on descending a group in the periodic table.
Answer:
Valence electrons.

Question 31.
An element X is in group 13 of the periodic table. What is the formula of its oxide?
Answer:
X₂O₃.

Question 32.
Elements ‘X’ and TP belong to groups 1 and 17 of the periodic table respectively. What will be the nature of the bond in the compound XY?
Answer:
Ionic bond.

Extra Questions for Class 10 Science Chapter 5 Short Answer Type I

Question 1.
The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called? Give one example of such a set of elements.   [NCERT Exemplar]
Answer:
The arrangement of these elements is known as Dobereiner triad. Example, Lithium, sodium and Potassium.

Question 2.
In group I of the periodic table, three elements X, Y and Z have atomic radii 1.33 A, 0.95 A and 0.60 A respectively. Arrange the elements (X, Y and Z) in the increasing order of atomic number and mention a suitable reason for it.
Answer:
In a given group, the atomic radii increases on moving down the group. Therefore, the arrangement of the given elements in the increasing order of their atomic numbers is as follows: Z, Y, X.

Question 3.
Elements have been arranged in the following sequence on the basis of their increasing atomic masses.
F, Na, Mg, Al, Si, P, S, Cl, Ar, K
(a) Pick two sets of elements which have similar properties.
(b) The given sequence represents which law of classification of elements?   [NCERT Exemplar]
Answer:
(a) F and Cl are first and eighth element in the above sequence, therefore, they have similar properties.
Please note that although Na and K have similar properties but they are not related as first and eighth element in the above sequence.

(b) This sequence represents Newland’s Law of Octaves.

Question 4.
State and explain Mendeleev’s periodic law.
Answer:
According to Mendeleev’s periodic law “Physical and chemical properties of elements are periodic function of their atomic masses”.

If elements are arranged in the increasing order of atomic masses, elements with similar properties lies after a certain interval in the periodic table.

Question 5.
“Hydrogen occupies a unique position in the Modern Periodic Table”. Justify the statement.   [NCERT Exemplar]
Answer:
The position of the element hydrogen is still not clear even in the Modern Periodic Table.

In electronic configuration, it resembles alkali metals of group 1. All of them have only one electron in the valence shell. Actually, hydrogen has only one shell (K-shell) which has one electron.

In characteristics, it resembles halogens of group 17. For example, like halogens hydrogen is a non-metal and diatomic as well. It has been therefore, decided to assign hydrogen a unique position in the Modern or Long Form Periodic Table. It is placed at the top in group 1 of alkali metals. However, it is not a member of that group.

Question 6.
If an element X is placed in group 14, what will be the formula and the nature of bonding of its chloride?  [NCERT Exemplar]
Answer:
Since element “X’ is placed in group 14, therefore, its valency is 14 – 10 = 4. Further, since it is difficult to either lose all the four valence electrons or gain four more electrons, therefore, it prefers to share these four electrons to acquire the stable electronic configuration of the nearest inert gas. Its formula will be XCl₄. Thus, the nature of the chloride of element ‘X’ is covalent.

Question 7.
Explain clearly, why atomic number of an element is most important to the chemist than its relative atomic mass.
Answer:
Atomic number corresponds to the number of electron in an atom or it reflects the electronic configuration of the element. The elements having similar electronic configurations can then be placed together in the same group, it helps in the systematic classification of elements.

On the other hand, the atomic mass of an element does not provide the electronic configuration of an element, so atomic number is more fundamental property for classification of elements.

Question 8.
What is the need of classification of the elements?
Answer:
The following reasons can be assigned for practical utility of classification of elements:

• To make the study of chemical elements easier.
• To correct atomic masses of various elements.
• To discover new elements.

Question 9.
The elements calcium, strontium and barium were put in one group or family on the basis of their similar properties.
(i) Mention the two similar properties?
(ii) What is the usual name of this group or family?
Answer:
(i) (a) All these elements are metals.
(b) All these elements have a valency of 2.

(ii) The usual name of this group or family is alkaline earth metals.

Question 10.
How does the electronic configuration of an atom of an element relate to its position in the modem periodic table? Explain with one example.   [CBSE 2011 (Delhi)]
Answer:
The electronic configuration of an atom of an element gives its position in the modern periodic table.
(i) The ‘period number’ of an element is equal to the number of electron shells in its atom.
(ii)

• The group number of an element having upto two valence electrons is equal to the number of valence electrons.
• The group number of an element having more than 2 valence electrons is equal to the number of valence electrons plus 10.

Example: If the electronic configuration of an element is 2, 8, 7.
Then its period number is 3 as it has three electrons shells.
Its group number is 17 as it has 7 valence electrons. (∵ Group no. = 7 + 10 = 17)

Question 11.
An element X (atomic number 17) combines with an element Y (atomic number 20) to form a compound.
(i) Write the positions of these elements in the modern periodic table.
(ii) Write the formula of the compound formed. Justify your answer in each case.   [CBSE 2013]
Answer:
(i) The electronic configurations of the two elements are:
X (Z = 17) 2, 8, 7 ; Y (Z = 20) 2, 8, 8, 2.
Element X is present in group 17 (halogen family) since it has 7 valence electrons. Element Y is placed in group 2 (alkaline earth family) since it has 2 valence electrons.

(ii) Element Y is a metal with valency 2 while element X is a non-metal with valency 1. Therefore, they two combine to form compound YX₂.

Question 12.
Explain why:
(i) All the elements in a group have similar chemical properties.
(ii) All the elements in a period have different chemical properties.
Answer:
(i) All the elements in a particular group have similar outer shell electronic configuration. Since chemical properties of an element are determined mainly by the outer shell configuration, all the elements in a group have similar chemical properties.

(ii) All the elements in a period have different valence shell electronic configuration because they have different number of electrons in the valence shell. Hence, the elements in a period have different chemical properties.

Question 13.
Given below are the atomic radii of three elements X, Y and Z of the periodic table, each having n electrons in the outermost shell of their atoms:

Element	X	Y	Z
Atomic radii	133 pm	157 pm	202 pm

Answer the following:
(a) Do these elements X, Y and Z belong to same group or to the same period?
(b) Which element will be least metallic?
Answer:
(a) Since the elements X, Y and Z contain same number of electrons (n) in the valence shell (outermost shell), they belong to the same group.
(b) In a group, the atomic radius increases on descending the group. Therefore, the element X with the smallest atomic radius is at the top of the group and the element Z having largest atomic radius is at the bottom.

Now, we know metallic character increases as we move from top to bottom in a group. Therefore, the least metallic element is X.

Question 14.
(a) The elements of the second period along with their atomic numbers in parentheses are given below:
B (5), Be (4), O (8), N (7), Li (3), C (6), F (9)
(i) Arrange them in the same order as they appear in the periodic table.
(ii) Which element has the largest and smallest atom?
(b) Why does atomic radius change as we move from left to right in a period?   [CBSE 2011]
Answer:
(a) (i) Li (3), Be (4), B (5), C (6), N (7), O (8), F (9)
(ii) The element Li has the largest atom. The element F has the smallest atom.

(b) Along a period, the nuclear charge increases and the electrons are attracted more towards the nucleus. Therefore, the atomic size or atomic radius decreases as we move from left to the right along a period.

Question 15.
Write two reasons responsible for the late discovery of noble gases.   [CBSE 2013]
Answer:

1. Noble gas elements were not present in earth crust as minerals like other elements and were present in air to a very small extent.
2. Their atoms have stable electronic configuration of their outermost shells also called valence shells. (2 in case of He and 8 for other elements). They do not combine with atoms of other elements.

That is why, noble gas elements were discovered at a later stage.

Question 16.
From the following elements:
₄Be; ₉F; ₁₉K; ₂₀Ca
(i) Select the element having one electron in the outermost shell.
(ii) two elements of the same group.
Write the formula of and mention the nature of the compound formed by the union of ₁₉K and element X (2, 8, 7). [Delhi 2015]
Answer:
(i) ₁₉K
(ii) ₄Be, ₂₀Ca belong to the same group.
Electronic configuration of K (19) = 2, 8, 8, 1   Valency of K = 1
Electronic configuration of X (2, 8, 7)   Valency of X = 1

The formula of the compound formed is KX. The compound KX is of ionic nature.
The bond is formed by transference of electrons.

Question 17.
Na, Mg and Al are the elements of the same period of Modern Periodic Table having one, two and three valence electrons respectively. Which of these elements (i) has the largest atomic radius, (ii) is least reactive? Justify your answer stating reason for each case.   [Delhi 2015]
Answer:

(1) Na has the largest atomic radius because on moving from left to right in the periodic table, the atomic radius decreases due to increase in positive charge on the nucleus which pulls the outermost electrons more close to the nucleus and the size of atom decreases.

(2) Al is least reactive because on moving from left to right in the periodic table the nuclear charge increases and the valence electrons are pulled in more close to the nucleus. Therefore, the tendency to lose electrons decreases and hence reactivity decreases.

Question 18.
How many groups and periods are there in the modern periodic table? How do the atomic size and metallic character of elements vary as we move: (a) down a group and (b) from left to right in a period. [Delhi 2015]
Answer:
Number of groups is 18 and number of periods in the modern periodic table is 7.
(a) All the elements in a group have the same valency. On going down in a group, the atomic size increases because a new shell of electrons is added to the atoms at every step.

(b) On moving from left to right in the periodic table, the valency of the elements first increases from 1 to 4 and then decreases to zero. On moving from left to right in a period, the metallic character of elements decrease because on moving from left to right in a period, the electropositive character of elements decrease.

Question 19.
The electronic configuration of an element X is:

(i) What is the group number of element X in the periodic table?
(ii) What is the period number of element X in the periodic table?
(iii) What is the number of valence electrons in an atom of X?
(iv) What is the valency of X?
Answer:
(i) From the above given electronic configuration we find that element X has 6 valence electrons in the outermost shell), so the group number of element X in the periodic table is 6 + 10 = 16.

(ii) Element X has 3 electron shells (K, L and M) in its atom, so the period number of X is 3. That is, X belongs to 3rd period of the periodic table.

(iii) Element X has 6 valence electrons.

(iv) Element X has 6 valence electrons so it needs 2 more electrons to complete its octet (8 electrons in valence shell) and become stable. Thus, the valency of element X is 2.

Extra Questions for Class 10 Science Chapter 5 Short Answer Type II

Question 1.
Identify the elements with the following property and arrange them in increasing order of their reactivity.
(a) An element which is a soft and reactive metal.
(b) The metal which is an important constituent of limestone.
(c) The metal which exists in liquid state at room temperature.  [NCERT Exemplar]
Answer:
(a) Alkali metals are soft and reactive. For example, Na, K, etc.
(b) Limestone is calcium carbonate, therefore, the important constituent of limestone is calcium.
(c) Metals which exists in the liquid state at room temperature are Mercury (Hg).

Question 2.
List three main anomalies of Mendeleev’s classification of elements.
Answer:
Anomalies or defects of Mendeleev’s periodic classification of elements are as follows:

1. Position of hydrogen.
2. Position of isotopes: Isotopes of an element were not assigned separate places.
3. Elements which are chemically similar such as gold and platinum have been placed in separate groups.

Question 3.
(a) State modern periodic law.
(b) What are the advantages of the long form of the periodic table over Mendeleev’s periodic table?
Answer:
(a) According to modern periodic law “physical and chemical properties of elements are periodic function of their atomic numbers”.

(b) (i) Classification of elements on the basis of atomic number is more closer to chemical properties.
(ii) No separate places for isotopes of an element are required.
(iii) In case of Ar and K, Ar has less atomic number so it should be placed before K according to the increasing order of atomic number.

Question 4.
Two elements X and Y have atomic number 12 and 17 respectively.
(i) Write the electronic configuration of both.
(ii) Which type of bond will they form?
(iii) Write the formula of the compound formed by their combination (in terms of X and Y).
Answer:
(i) The electronic configuration of element X (Z = 12) = 2, 8, 2
The electronic configuration of element Y (Z = 17) = 2, 8, 7

(ii) The bond formed will be of ionic nature. One atom of X will transfer 1 electron each to two atoms of Y.

(iii) The formula of the compound is:

Question 5.
Name:
(a) Three elements that have a single electron in their outermost shells.
(b) Two elements that have two electrons in their outermost shells.
(c) Three elements with filled outermost shells.
Answer:
(a) Lithium (Li), Sodium (Na), Potassium (K)
(b) Magnesium (Mg), Calcium (Ca)
(c) Helium (He), Neon (Ne), Argon (Ar)

Question 6.
Two elements X and Y belong to group 1 and 2 respectively in the same period of periodic table. Compare them with respect to:   [CBSE 2011]
(i) the number of valence electrons in their atoms
(ii) their valencies
(iii) metallic character
(iv) the sizes of their atoms
(v) the formulae of their oxides
(vi) the formulae of their chlorides.
Answer:
X and Y belong to same period.

• X belongs to group ‘1’.
• Y belongs to group ‘2’.

(i) Valence electron in X is 1 whereas valence electrons in Y are 2.
(ii) The valency of X is 1 whereas valency of Y is 2.
(iii) X is more metallic than Y because metallic character decreases on moving from left to right in a period.
(iv) The size of X is more than Y because size of the atom decreases on moving from left to right in a period.
(v) Oxide of X = X₂O,   Oxide of Y = YO
(vi) Chloride of X = XCl,   Chloride of Y = YCl₂

Question 7.
The element Li, Na and K, each having one valence electron, are in period 2, 3 and 4 respectively of the Modem Periodic Table.
(a) In which group of the periodic table should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic radius? Give reason to justify your answer in each case. [CBSE 2013]
Answer:
(a) Since the elements have one valence electron, they are placed in group 1 (Alkali metals) in the order Li (period = 2); Na (period = 3); K (period = 4).

(b) Since the reactivity of the elements increases down a group, the element Li is the least reactive chemically.

(c) Since the atomic size increases down a group, the element K has the large atomic size or atomic radius out of these elements.

Question 8.
Explain why are the following statements not correct:
(a) All groups contain metals and non-metals.
(b) Atoms of elements in the same group have the same number of electron(s).
(c) Non-metallic character decreases across a period with increase in atomic number.
(d) Reactivity increases with atomic number in a group as well as in a period.   [2010]
Answer:
(a) Because all groups do not contain metals and non-metals. For example, Alkali metal group contains metals only.
(b) Atom of the elements in the same group have same number of electrons in the valence shell.
(c) Non-metallic character increases across a period with increase in atomic number.
(d) Reactivity increases with atomic number in a group but not across a period.

Question 9.
This question refers to the elements of the periodic table with atomic numbers from 3 to 18. Some of the elements are shown by letters but the letters are not the usual symbols of the elements.

(a) Which of these
(i) are noble gases?
(ii) are halogens?
(iii) are alkali metals?
(b) If A combines with F, what would be the formula of the resulting compound?
(c) What is the electronic configuration of G?
Answer:
(a) (i) H, P;
(ii) G, O;
(iii) A, I
(b) F
(c) 2, 7

Question 10.
Name two other elements which are in the same group as:
(i) Carbon
(ii) Fluorine
(iii) Sodium, respectively.
Answer:
(i) The two other elements which are in the same group as carbon are: Silicon (Si) and germanium (Ge).
(ii) The two other elements which are in the same group as fluorine are : Chlorine (Cl) and bromine (Br).
(iii) The two other elements which are in the same group as sodium are : Lithium (Li) and potassium (K).

Question 11.
Atomic number of an element is 16. Predict
(i) the number of valence electrons in its atom
(ii) its valency
(iii) its group number
(iv) whether it is a metal or a non-metal
(v) the nature of oxide formed by it.
(vi) the formula of the chloride   [CBSE2011]
Answer:
The element with atomic number Z = 16 is sulphur. The electronic configuration of the element is 2, 8, 6.
(i) The number of valence electrons is 6
(ii) The valency of the element is 2 (8 – 6 = 2)
(iii) The group number of the element is 16.
(iv) The element is a non-metal.
(v) Since the element is non-metal, its oxide is acidic in nature. Actually, two oxides are formed. These are SO₂ and SO₃. Both are of acidic nature.
(vi) The formula of the chloride is SCl₂.

Question 12.
(a) How are the following related?
(i) Number of valence electrons of the different elements present in the same group.
(ii) Number of the shells of the elements in the same period.
(b) How do the following change:
(i) Number of shells of the elements as we go down a group.
(ii) Number of valence electrons of elements on moving from left to right in a period.
(iii) Atomic radius in moving from left to right along a period.
(iv) Atomic size down a group. [CBSE 2012]
Answer:
(a) (i) Number of valence electrons of the elements present in a group do not change.
(ii) The elements present in the same period have same number of shells.

(b) (i) The number of shells of elements in a group gradually increase downwards.
(ii) The number of valence electrons of the elements in moving from left to right along a period gradually increase.
(iii) Atomic radius decreases from left to the right.
(iv) Atomic size increases down a group.

Question 13.
(a) What is meant by periodicity in properties of elements with reference to the periodic table?
(b) Why do all the elements of the same group have similar properties?
(c) How will the tendency to gain electrons change as we go from left to right across a period? Why?
Answer:
(a) Periodicity in properties of elements with reference to the periodic table means the elements having similar properties are repeated after certain intervals or periods and the elements are arranged in the tabular form.

(b) All the elements of the same group have similar properties because all the elements belonging to the same group of the Periodic Table have same number of valence electrons.

(c) On moving from left to right in a period, the tendency of atoms to gain electrons increases because on moving from left to right in a period the nuclear charge (positive charge on nucleus) increases due to gradual increase in the number of protons so it becomes easier for the atoms to gain electrons.

Question 14.
What is meant by ‘group’ in the modern periodic table? How do the following change on moving from top to bottom in a group?
(i) Number of valence electrons
(ii) Number of occupied shells
(iii) Size of atoms
(iv) Metallic character of elements
(v) Effective nuclear change experienced by valence electrons.  [CBSE 2014]
Answer:
The vertical columns in a periodic table are called groups.
(i) All the elements of a group have the same number of valence electrons.
(ii) On moving down in a group the number of occupied (filled) shells increases gradually.
(iii) On going down in a group the size of atoms increases because a new shell of electrons is added to the atoms at every step.
(iv) On going down in a group the metallic character of elements increases.
(v) On moving down in a group, one more electron shell is added at every stage and size of the atom increases. Thus valence electrons move more and more away from the nucleus and hold of the nucleus or nuclear charge on valence electrons decreases.

Question 15.
The elements Be, Mg and Ca each having two electrons in their outermost shells are in periods 2, 3, and 4 respectively of the modern periodic table. Answer the following questions, giving justification in each case:  [CBSE 2014]
(i) Write the group to which these elements belong.
(ii) Name the least reactive element.
(iii) Name the element having largest atomic radius.
Answer:

Element	Valence electrons	Period	Electronic configuration
Be	2	2	2,2
Mg	2	3	2, 8, 2
Ca	2	4	2, 8, 8, 2

(i) All these elements belong to the 2nd group all have two electrons in their outermost shell.

(ii) Be is the least reactive metal because reactivity of metals increases in a period as the tendency to lose electrons in a group increases.

Therefore, Be being the smallest in all the given elements of a period has its valance electrons nearest to the nucleus. So the removal of electrons from its valance shell will be difficult.

(iii) Ca has the largest atomic radius because it has maximum number of shells, i.e., 4.

Question 16.
An element ‘M’ with electronic configuration (2, 8, 2) combines separately with (NO₃)^–, (SO₄)^2- and (PO₄)^3- radicals. Write the formula of three compounds so formed. To which group and period of the Modern Periodic Table does the elements ‘M’ belong? Will ‘M’ from covalent or ionic compounds? Give reason to justify your answer.
Answer:
Element M with electronic configuration =

Total number of electrons = 2 + 8 + 2 = 12, i.e., the given element is Magnesium (Mg). Formula of compound formed are:

Formula of compound = Mg (NO₃)₂

Formula of compound = MgSO₄

Formula of compound = Mg₃ (PO₄)₂

M will form ionic compound with radicals given. An ionic compound is a chemical compound comprising ions held together by electrostatic forces. In this compound no sharing of electrons takes place.

Extra Questions for Class 10 Science Chapter 5 Long Answer Type

Question 1.
An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants.
(a) Identify the element X
(b) Write the electronic configuration of X
(c) Write the balanced chemical equation for the thermal decomposition of ferrous sulphate crystals.
(d) What would be the nature (acidic/basic) of oxides formed?
(e) Locate the position of the element in the Modern Periodic Table.  [NCERT Exemplar]
Answer:
(a) The element X which is a yellow solid at room temperature and shows catenation and allotropy is sulphur (S).

(b) The atomic number of sulphur is 16. Therefore, its electronic configuration is 2, 8, 6.

(c)

Both SO₂ and SO₃ are major air pollutants.

(d) Since sulphur is a non-metal, therefore, both SO₂ and SO₃ are acidic oxides since they dissolve in water to form the corresponding acids.

(e) Since sulphur contains six valence electrons, therefore, it lies in group 6 + 10 = 16. Further, since atomic number of sulphur (S) is 16, it lies in the 3rd period.

Question 2.
An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide.
(a) Where in the periodic table are elements X and Y placed?
(b) Classify X and Y as metal, non-metal or metalloid.
(c) What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed.
(d) Draw the electron dot structure of the divalent halide.   [NCERT Exemplar]
Answer:
(a) X belongs to Group 17 and 3^rd Period.
Y belongs to Group 2 and 4^th Period.
(b) X-non-metal and Y-metal.
(c) Basic oxide; Ionic bonding.
(d)

Question 3.
(a) Why do we classify elements?
(b) What were the two criteria used by Mendeleev in creating his Periodic Table?
(c) Why did Mendeleev leave some gaps in his Periodic Table?
(d) In Mendeleev’s Periodic Table, why was there no mention of Noble gases like Helium, Neon and Argon?
(e) Would you place the two isotopes of chlorine, Cl-35 and Cl-37 in different slots because of their different atomic masses or in the same slot because their chemical properties are the same? Justify your answer.
Answer;
(a) As different elements were being discovered, scientists gathered more information about the properties of these elements. It was observed that it was difficult to organise all the information or properties of these elements. So scientists started discovering some pattern in their properties to classify all the known elements to make their study easier.

(b) Atomic mass and similarity of chemical properties (compounds formed by elements with oxygen and hydrogen) were the two criteria used by Mendeleev in his Periodic Table.

(c) Mendeleev left some gaps in his Periodic Table as he predicted the existence of some elements that had not been discovered at that time.

(d) Noble gases like helium, neon, argon, etc. were not mentioned in Mendeleev’s Periodic Table because these gases were discovered later as they are very inert and present in extremely low concentrations in our atmosphere. After the discovery of noble gases they could be placed in a new group without disturbing the existing order of the Periodic Table.

(e) The Modem Periodic Table states that properties of elements are a periodic function of their atomic numbers, since Cl-35 and Cl-37 isotopes have the same atomic number (17), hence they will have same chemical properties even though their atomic masses are different. So, they should be placed in the same slot of the periodic table.

Question 4.
Atomic number of a few elements are give below:
10, 20, 7, 14
(a) Identify the elements.
(b) Identify the Group number of these elements in the Periodic Table.
(c) Identify the Periods of these elements in the Periodic Table.
(d) What would be the electronic configuration for each of these elements?
(e) Determine the valency of these elements.  [NCERT Exemplar]
Answer:
(a) The element are: Neon (Z = 10), Calcium (Z = 20), Nitrogen (Z = 7) and Silicon (Z = 14).
(b) Group number: Neon (18), Calcium (2), Nitrogen (15), Silicon (14).
(c) Periods: Neon (2), Calcium (4), Fluorine (2), Silicon (3).
(d) Electronic Configuration: Neon (2, 8); Calcium (2, 8, 8, 2); Nitrogen (2, 5); Silicon (2, 8, 4).
(e) Valency: Neon (zero); Calcium (2); Nitrogen (3); Silicon (4).

Question 5.
This question refers to the elements of the periodic table with atomic numbers from 3 to 18. Some of the elements are shown by letters, but the letters are not the usual symbols of the elements.

3	4	5	6	7	8	9	10
A					E		G
11	12	13	14	15	16	17	18
B	C		D			F

(a) Which of these:
(i) is a noble gas?
(ii) is a halogen?
(iii) is an alkali metal?
(iv) is an element with valency 4?
(b) If A combines with F, what would be the formula of the resulting compound?
(c) What is the electronic arrangement of G?
Answer:
(a) (i) G having 8 electrons in its valence shell is a noble gas.
₁₀G : 2, 8

(ii) F having 7 electrons in its valence shell is a halogen.
₁₇F : 2, 8, 7

(iii) B having 1 electron in its valence shell is an alkali metal.
₁₁B : 2, 8, 1
Element A is also an alkali metal.
₃A : 2, 1

(iv) The element D having 4 electrons in its valence shell will have valency 4.
₁₄D : 2, 8, 4

(b) A is an electropositive element with valency 1 while F is an electronegative element with valency 1.
Hence, the formula of the compound between A and F would be AF.

(c) The electronic configuration of G is:
₁₀G : 2, 8.

Question 6.
The elements of one short period of the periodic table are given below in the order from left to right.
Li, Be, B, C, O, F, Ne,
(a) To which period these elements belong?
(b) One element of this period is missing. Which is the missing element and where should it be placed?
(c) Which one of the element in this period shows the property of catenation?
(d) Place the three elements fluorine, beryllium and nitrogen in the order of increasing electronegativity.
(e) Which one of the above elements belong to the halogen series?
Answer:
(a) Second period (Atomic number: 3-10)
(b) Nitrogen (N), It should be placed between Carbon (Atomic number 6) and Oxygen (Atomic number 8)
(c) Carbon
(d) Be, N, F
(e) Fluorine.

Question 7.
Which element has
(a) two shells, both of which are completely filled with electrons?
(b) the electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?
Answer:
(a) Neon [Ne] (2, 8)
(b) Magnesium [Mg]
(c) Silicon [Si] (2, 8, 4)
(d) Boron [B] (2, 3)
(e) Carbon [C] (2, 4).

Periodic Classification of Elements HOTS Questions With Answers

Question 1.
(a) In this ladder symbols of elements are jumbled up. Rearrange these symbols of elements in the increasing order of their atomic number in the Periodic Table.
(b) Arrange them in the order of their group also.  [NCERT Exemplar]

Answer:
(a) H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca
(b) Group 1 – H, Li, Na, K
Group 2 – Be, Mg, Ca
Group 13 – B, Al
Group 14 – C, Si
Group 15 – N, P
Group 16 – O, S
Group 17 – F, Cl
Group 18 – He, Ne, Ar.

Question 2.
The electronic configuration of an element T is 2, 8, 7.
(а) What is the group number of T?
(b) What is the period number of T?
(c) How many valence electrons are there in an atom of T?
(d) Is it a metal or non-metal?
Answer:
(a) Group-17
(b) Third period
(c) 7
(d) Non-metal.

Question 3.
Mendeleev predicted the existence of certain elements not known at that time and named two of them as Eka-silicon and Eka-aluminium.
(a) Name the elements which have taken the place of these elements.
(b) Mention the group and the period of these elements in the Modern Periodic Table.
(c) Classify these elements as metals, non-metals or metalloids.
(d) How many valence electrons are present in each one of them?  [NCERT Exemplar]
Answer:
(a) Eka-silicon for germanium (Ge) and Eka-aluminium for Gallium (Ga).
(b) Group number of Ga is 13 and its period is 4^th and group number of Ge is 14 and its period is also 4^th.
(c) Both Ga and Ge are metalloids.
(d) Ga lies in group 13, therefore, it has 13 – 10 = 3 valence electrons. Similarly, Ge lies in group 14 and hence it has 14 – 10 = 4 valence electron.

Question 4.
The atomic numbers of three elements A, B and C are given below:

Element	Atomic number
A	5
B	7
C	10

(i) Which element belongs to group-18?
(ii) Which element belongs to group-5?
(iii) Which element belongs to group-3?
(iv) To which period/periods do these elements belong?
Answer:
(i) C
(ii) B
(iii) A
(iv) Second period.

Question 5.
(a) Electropositive nature of the element(s) increases down the group and decreases across the period.
(b) Electronegativity of the element decreases down the group and increases across the period.
(c) Atomic size increases down the group and decreases across a period (left to right).
(d) Metallic character increases down the group and decreases across a period.
On the basis of the above trends of the Periodic Table, answer the following about the elements with atomic numbers 3 to 9.
(a) Name the most electropositive element among them
(b) Name the most electronegative element
(c) Name the element with smallest atomic size
(d) Name the element which is a metalloid
(e) Name the element which shows maximum valency.  [NCERT Exemplar]
Answer:
(a) Lithium (Z = 3) is the most electropositive element.
(b) Fluorine (Z = 9) is the most electronegative element.
(c) Fluorine (Z = 9) has the smallest atomic size.
(d) Boron (Z = 5) is a metalloid.
(e) Carbon (Z = 6) shows the maximum valency (4).
However, the element nitrogen (Z = 7) can show valency 5 in some compounds (example, N₂O₅).

Question 6.
The following table shows the position of six elements A, B, C, D, E and F in the period table.

Groups/periods	1	2	3 to 12	13	14	15	16	17	18
2.		A				B			C
3.		D			E				F

Using the above table answer the following questions:
(a) Which element will form only covalent compounds?
(b) Which element is a metal with valency of 2?
(c ) Which element is a non-metal with valency of 3?
(d) Out of D and E, which one has more atomic radius and why?
(e) Write a common name for the family of elements C and F.
Answer:
(a) The element ‘E’ present in group 14 is a non-metal. Its name is silicon (Si) and the compounds of the element are only covalent.
(b) The element ‘D’ present in group 2 is a metal known as magnesium (Mg). It exhibits valency 2 in its compounds.(c) The elements ‘B’ present in group 15 is a non-metal. It is nitrogen (N) and exhibits valency 3 in its compounds.
(d) The element ‘D’ has more atomic radius than the element ‘E’ as the atomic size decreases along a period.
(e) The elements ‘C’ and ‘F’ present in group 18 belong to a family known as noble gases.

Question 7.
From the part of the periodic table given, answer the following questions

1 Lithium	2	13	14 Carbon	15	16 Oxygen	17 L	18 Neon
X			S		P	Q
Y						R
Z						T

(a) Which is the most reactive metal?
(b) Name the family of L, Q, R and T.
(c) Name one element of group 2 and 15.
(d) Name one member of group 18 other than neon.
(e) Give the name of the element S placed below carbon in group 14.
Answer:
(a) Z
(b) L, Q, R and T belongs to halogen family.
(c) Group-2 element – Ca, Mg
Group-15 element – N, P
(d) The element argon (Ar) is also present in group 18.
(e) The element is silicon (Si).

Extra Questions for Class 10 Science Chapter 5 Value Based Questions

Question 1.
First systematic classification was done by Mendeleev. He arranged the elements in the increasing order of atomic masses. Elements with similar properties lies in a group, he left certain gaps for unknown elements.
(i) Name two elements for which blank spaces were left out by Mendeleev.
(ii) List two main drawbacks of the Mendeleev’s periodic classification.
(iii) Mention the values exhibited by Mendeleev.
Answer:
(i) Gallium (Ga), Germanium (Ge).
(ii) (a) Position of hydrogen.
(b) Position of isotopes.
(iii) Use of knowledge of Chemistry, Thinking ability.

Question 2.
Ria and Ram are students of class X. Ria is very organised and systematic. The teachers love her. She earns a great respect in the class whereas Ram is unorganised and always faces a lot of problem in handling life situations.
(i) In your opinion how does the organisation help in daily life?
(ii) How can you relate the above fact with the chapter classification of elements. How does classification of elements help us in studying them properly?
Answer:
(i) Organised approach always make the work easier and quicker. Hence, classification makes the things easier.

(ii) Organised approach in classification of elements make the study easier and we get a group of elements with similar properties.
By this approach we can study the elements easily which lies in a group with similar chemical properties.

Here we are providing Class 10 History Chapter 3 Extra Questions and Answers Nationalism in India was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-social-science/

Nationalism in India Class 10 Extra Questions History Chapter 3

Question 1.
Mention the significance of the Home Rule Movement. State its two features.
Answer:
The significance of the Home Rule movement in India’s freedom struggle cannot be denied. It gave the nationalist movement a boost needed at that time. The two features of the Home Rule Movement were:

1. Demand for a dominion status.
2. Demand for Home rule.

Question 2.
What do you mean by the Kakori Conspiracy Case? Name the two terrorists hanged in this case.
Answer:
The British government struck at the workers of the Hindustan Republican Association by arresting a large number of terrorist youth and trying them in Kakori Conspiracy Case (1925) Ram Prasad Bismil and Ashfaqulla were hanged in this case.

Question 3.
Why did the British government appoint Simon Commission? Why did the Indians boycott it?
Answer:
The British government appointed the Simon Commission” in 1923 to examine the working of the Government of India Act, 1919.
The Indians boycotted the Commission because,

• it was appointed by the Conservative Party government, unsympathetic towards India’s cause,
• it consisted of all Englishmen.

Question 4.
What were the reasons for launching Civil Disobedience Movement by the Congress?
Answer:

• The Indian National Congress had already resolved in the Lahore session of 1929 to launch the movement.
• The British Government was unwilling to solve the Indian”political problems, i.e., liberation of the country.

Question 5.
What were the two important decisions taken in the Lahore session of Indian National Congress of 1929?
Answer:
A session of the Indian National Congress was held at Lahore in 1929. The two important decisions taken here were :

1. Declaration of complete independence and its unilateral celebration on January 26, 1930.
2. Launching of a civil disobedience movement in 1930.

Question 6.
What were the objectives of the Constituent Assembly (1946)?
Answer:
The following were the objectives of the Constituent Assembly :

• To make the Constitution of India.
• To act as the Parliament of India When the British leave the country.

Question 7.
What do you know about the Montague’s Declaration?
Answer:
The Montague’s declaration was made in August 1917. The British policy of administration was pronounced. It was said that the ultimate aim of the British rule in India was the establishment of responsible government in the country.

Question 8.
What was dyarchy in the context of the Government of India Act, 1919?
Answer:
Dyarchy, in the context of the Government of India Act, (1919) was the introduction of the dual rule in the provinces: the reserved departments under the councillors, all Englishmen; the transferred departments under the ministers, all Indians.

Question 9.
Why is Dandi March important? Give reason.
Answer:
Dandi March is important because it was at that place” that Gandhiji made all out of the seawater on April 6, 1930. It was the beginning of the Civil Disobedience Movement.

Question 10.
When and by whom was the Indian National Army (IN A) reorganised? What was its main purpose?
Answer:
The Indian National Army was reorganised by Subhash Chandra Bose in Singapore in 1943. The main purpose of INA was to liberate India from the yoke of the British rule.

Question 11.
Who was the architect of the two-nation theory? What was its main object?
Answer:
The two-nation theory was advocated by M.A.Jinnah The main purpose of this theory was to have two separate independent states for the Hindus and the Muslims.

Question 12.
Why did Gandhiji launch individual satyagraha? Why was the first satyagraha?
Answer:
Gandhiji launched individual Satyagraha in 1940 with the object of dissuading the Indians in joining the war. Vinoba Bhave was the first satyagrahi.

Question 13.
How did the people discover their unity while fighting”against colonialism?
Answer:
People began discovering their unity in the process of their struggle against colonialism.

Question 14.
Why did forced recruitment cause widespread anger India on the eve of world war?
Answer:
On the eve of World War I, the villagers in India faced hardships such as inflation, increasing custom duties, taxation sailing crops. When the British government began forced recruitment in the army, the villagers were angry.

Question 15.
When was the Khilafat committee formed and for what purposes?
Answer:
The Khilafat Committee was formed in Bombay in 1919 to defend the temporal powers of the Khalifa, the spiritual head of the Islamic world.

Question 16.
Why was the Nagpur session of the Congress important?
Answer:
The Nagpur session of the Congress (1920) was important because it was in this session that the non-cooperation programme was adopted.

Question 17.
Why was the rural India in turmoil by 1930?
Answer:
The economic depression had its effect all over the world. Agricultural prices began to fall from 1926 and collapsed after 1930. The demand for agricultural products also fell leading to the decline of export. The peasants could not sell their products, nor could they pay their revenue.

Question 18.
Why was the Lahore session of the Congress (1929) important?
Answer:
The Lahore session of the Congress (1929) was important because it was here that the Congress passed the “Purna Swaraj” resolution. The Indians had celebrated 26th January 1930 as the independence day.

Question 19.
Mention the social groups which were lukewarm to the civil disobedience movement.
Answer:
The Dalits, the upper-caste Hindus, some Muslim organisations were lukewarm to the civil disobedience movement launched by Gandhiji.

Question 20.
It is said that the ideas of nationalism developed through a movement to revive Indian folklore? Give example.
Answer:
It is true that the ideas of nationalism developed through a movement to revive Indian folklore. In Bengal, Rabindranath Tagore, for example, himself began collecting ballads, nursery rhymes and myths and led the movement for folk revival.

Question 21.
When and why was the Muslim League founded? What was the main idea which led to I the partition of India in 1987 Elucidate?
Answer:
The Muslim League was founded in 1906 at Dhaka. Its”objectives included the following:

• To protect and promote the interests of the Muslims.
• To bring the Muslims and the British closer in each other.
• To foster friendship with other communities of India.

The main idea which led to the partition of India in 1947 was Jinnah’s two-nation theory and his direct action strategy (use of violence) to get Pakistan.

Question 22.
What was the contribution of the revolutionaries to the growth of Indian Nationalism?
Answer:
The contribution of the revolutionaries to the growth of Indian nationalism canot be underestimated. They awakened the common masses and filled in them the spirit of nationalism. They attempted to highlight injustice metal out The Indians in India as well as abroad. They demanded supreme sacrifice from the people to get the country free from the British.

Question 23.
Describe the role of revolutionaries in the Indian National Movement.
Answer:
The role of the revolutionaries in the Indian National Movement can hardly be denied. The revolutionaries had no faith in the British and hence wanted them to leave the country. For ousting the British, the revolutionaries did not hesitate in resorting to the extremist and revolutionary activities. In Punjab, Bengal, Maharashtra and Chennai (Madras), they followed policy of terrorising the British rulers. They were successful in generating”awakening among the people.

Question 24.
Discuss four important provisions of the Act of 1919.
Answer:

1. The Act of 1919 increased the membership strength of the legislative bodies: both at the centre as well as in the states.
2. It enhanced the powers of the legislative bodies.
3. At the provincial level, dyarchy was introduced.
4. The Governor-General and the Governors got discretionary powers to override the decisions of the legislative bodies.

Question 25.
What was the Rowlatt Act? Why did the Indians oppose it?
Answer:
The Rowlatt Act was an Act passed by the British Government in India. The Act authorised the police to arrest and imprison any person without trying him in any court. It was an oppressive Act. The Indians opposed the Ad because it was detrimental to the freedoms and liberties of the people. As it was also arbitrary and dictatorial; the Act was condemned by the’Tndians.

Question 26.
Why and by whom was the Khilafat Movement organised?
Answer:
The Khilafat movement was organised by Ali brothers Shaukat Ali and Mohammad Ali. They launched this movement

• to prevent the possibilities of dis membership of Turkey following its defeat in the First World War,
• to ask the Muslims not to join the army on the side of the English,
• to strengthen Hindu Muslim unity.
• to establish the glory of the Turkish Ottoman Empire.

Question 27.
Analyse the significance of the Khilafat and Non-Cooperation Movement in India’s struggle for freedom.
Answer:
The Khilafat and the Non-Cooperation Movements had their special role in the history of India’s freedom movement. The Khilafat movement was launched by Ali brothers and the Non-Cooperation Movement, by Mahatma Gandhi The leaders of both the movements joined each other’s movement and shook the very foundations of the British Government. Both the movements laid stress on the Hindu Muslim unity, both brought the mass of the people in liberation struggle against the British imperialism.

Question 28.
Why did Gandhiji withdraw the Non-Cooperation Movement?
Answer:
Gandhiji had launched Non-Cooperation Movement in 1920. The movement went so well that the Britishers were shaken. The lawyers came out of the courts, the students from the schools and colleges. The Indians boycotted official functions. All of a sudden, there broke out violence in a place called Chauri Chaura where the mob had attacked a police station killing 22 police personnel. As Gandhiji wanted the movement to be non-violent the use of violence forced him to withdraw the movement.

Question 29.
What was the significance of 1931 Karachi session of the Congress.
Answer:
Significant contribution of 1931 Karachi session:

• The Congress session at Karachi in 1981 approved the Gandhi-Irwin Pact.
• The most significant contribution of the Karachi session was a resolution passed on Fundamental Rights and Economic policy
• The Congress suspended the civil disobedience movement and agreed to take part in Round Table Conference in London.

Question 30.
When and why did the Quit India Movement start? What were its effects on the national movement?
Answer:
The Quit India movement was launched in August. 1942. It was launched so to ask the British quit the country, for their failure to grant independence to India. As the Indian leaders were arrested before launching of the movement, the movement came to be launched by the people.
The while country rose against the British. Despite the repressive measures of the British Government, the movement spread in the whole country. It shook the British rule.

Question 31.
In what way did communalism obstruct the growth of nationalism in India?
Answer:
Communalism was a potent factor which had obstructed the growth of nationalism in India. It did so because it worked in the framework of the narrow interests. It separated one set of people from the other set of people. It created wide distances between the people belonging to one religion and those belonging to the other. It resulted in enmity and hatred among the different communities and led to communal riots. The communal enmity had ultimately ended in the partition of India into two dominions.

Question 32.
What was the role played by Azad Hind Fauj (Indian National Army) in India’s struggle for freedom?
Answer:
The role played by the Indian National Army in India’s struggle for freedom cannot be denied. In fact, the Indian National Army proved a serious external danger to the British. It forced the British to leave the country as early as possible. Subhash Chandra Bose gave the slogans of ‘Delhi Chalo’ and ‘Jai Hind’, to the Indians.

Question 33.
State briefly the effects of non cooperation on the economic front.
Answer:
The effects of non-cooperation on the economic front were more dramatic Foreign goods were boycotted, liquor shops picketed. and foreign cloth burnt in huge bonfires. The import of foreign cloth halved between 1921 and 1922, its value dropping from ₹ 102 crore to ₹ 57 crore. In many places, merchants and traders refused to trade in foreign goods or finance foreign trade.

As the boycott movement spread, and people began discarding imported clothes and wearing only Indian ones, production of Indian textile mills and handlooms went up.

Question 34.
Explain as to why the non-cooperation movement slowed down in the cities. Give reasons. ‘
Answer:
The non-cooperation movement in the cities gradually slowed down for a variety of reasons. Khadi cloth was ofter more expensive than mass-produced mill cloth and poor people could not afford to buy it. How then could they boycott mill cloth for too long? Similarly, the boycott of British institutions posed a problem.

For the movement to be successful, alternative Indian institutions had to be set up so that they could be used in place of the British ones. These were slow to come up. So students and teachers began trickling back to government schools and lawyers joined back work in government courts.

Question 35.
Who was Baba Ramchandra ?
Answer:
Baba Ramchandra was a sanyasi who led the peasant movement in Awadh. The movement was against talukdars and landlords who demand high rents from peasants. Peasants had to do begar and work at landlords farms without any payment. As tenants they had no security of tenure, being regularly evicted so that they could acquire no right over the leased land.

The peasant movement demanded reduction of revenue, abolition of legal, and social boycott of oppressive landlords. As the movement spread in 1921, the houses of talukdars and merchants were attacked bazaars were looted, and grain hoards were taken over.

Question 36.
“Nationalism, in the early 20 century, came to be associated with the image of Bharat Mata”. Explain with illustration.
Answer:
It was in the twentieth century, with the growth of nationalism, that the identity of India came to be visually associated with the image of Bharat Mata. The image was first created by Bankim Chandra Chattopadhyay. In the 1870s he wrote ‘Vande Mataram’ as a hymn to the motherland.

Later it was included in his novel Anandamath and widely sung during the Swadeshi movement in Bengal. Moved by the Swadeshi movement, Abanindranath Tagore painted his famous image of Bharat Mata. In this painting Bharat Mata is portrayed as an ascetic figure; she is calm, composed, divine and spiritual.

Question 37.
How did the tribal peasants interpreted the message of Gandhiji and the satyagraha? Explain it with the Gudem revolt.
Answer:
Tribal peasants interpreted the message of Mahatma”Gandhi and the idea of Swraj in yet another way. In the Gudem Hills of Andhra Pradesh, for instance, a militant guerrilla movement spread in the early 1920s – not a form of struggle that the Congress could approve. Here, as in other forest regions, the colonial government had closed large forest areas, preventing people from entering the forests to graze their cattle, or to collect fuelwood and emits.

This enraged the hill people. Not only were their livelihoods affected but they felt that their traditional rights were being denied When the government began forcing there to”contribute begar for road building, the hill people revolted. The person who came to lead them was an interesting figure. Alluri Sitaram Raju claimed that he had a variety of special powers. He could make correct astrological predictions and heal people, and he could survive even bullet shots.

Captivated by Raju, the rebels proclaimed that he was an incarnation of God. Raju talked of the greatness of Mahatma Gandhi, said he was inspired by the Non-Cooperation Movement and persuaded people to wear khadi and give up drinking But at the same time he asserted that India could be liberated only by the use of force, not non-violence. The Gudem rebels attacked police stations, attempted to kill British officials and carried on guerrilla warfare for achieving swaraj. Raju was captured and executed in 1924, and over time became a folk hero.

Question 38.
What were the Khilafat and the Non- cooperation” movements? Name some important leaders of these movements.
Answer:
Khilafat Movement: It was a movement launched by Ali brothers-Shaukat All and Mohammad Ali to oppose the British policy of dismembering the Turkish Ottoman Empire and the office of the Caliph in Turkey, the religious head of the Muslim community.

Non-Cooperation Movement: It was a movement launched by Mahatma Gandhi in 1920 to protest against the British policy of repression through the Rowiati Act and the Jallianwala massacre in 1919.

Gandhiji launched the movement to demonstrate non-cooperation with the British Government by boycotting elections, schools and courts on the one hand, and on the other, attempting to bring about unity among the various sections of the Indian society by using Charkha and Swadeshi The movement was to be carried on through peaceful measures. But an accident occurred at Chauri-Chaura. The movement was withdrawn in 1922.

The leaders of the Khilafat movement were Shaukat Ali and Mohammad Ali. The leader of the Non-Cooperation Movement was Mahatma Gandhi.

Question 39.
What did the slogan of Swaraj mean? How was the slogan of Complete Independence different from it? When and where was the slogan of Complete Independence adopted?
Answer:
Swaraj’ meant that the autonomy be given to the Indians. Following the Calcutta (now Kolkata) Congress session in 1905, the Indian. National Congress sought to demand Swaraj. What Swaraj meant was the freedom to rule ! which was to mean that the power to administer the affairs be handed over to the Indians.

In other words, it meant that the power to be given to the elected Indians who should be responsible to the legislature, and the legislature to be responsible to the people to be more specific, the rulers have to be Indians, and not the English.

Complete independence or what may be called the Purna Swaraj was to be different from the word ‘Swaraj’. Swaraj had the indication of internal autonomy, though the external autonomy was to be exercised by the British The Purna Swaraj or the complete independence, as a resolution was passed at the Lahore Congress Session of 1929.

Through the 1929 complete independence resolution, the Congress sought complete freedom from the British internal autonomy as well as freedom to act on its own in relation to the other countries. The complete independence resolution was passed by the Congress in 1929 at the Lahore Congress Session.

Question 40.
When was the Muslim League formed? Describe briefly the policies of the Muslim League from 1906 to 1940. When was the formation of a separate state of Pakistan adopted as the aim of the Muslim League?
Answer:
The Muslim League was formed in December 1906 with the aim of seeking the Muslim representation in the government and bringing the Britishers and the Muslims, closer to each other, From the day one, the League thought of itself, though mistakenly, as the sole representative of the Muslims, and began projecting the Congress as the Flindu organisation.

From 1906 to around 1911, the Britishers supported the League and the league approved of the British policies in India. The League supported the proposal of the Bengal partition: it did not participate in the nationalist movement against the proposals of Bengal partition. But as the war aims of the British home Government against the Turkish Empire, dismembering it became clearer, the League and the British Government in India drew distant and consequently the League drew closer to the Congress.

The League and the Congress annual sessions began at the same time and in the same cities in 1914, 1915, 1916. There was the Lucknow Pact (1916) between the Congress and the League: the Congress agreeing the communal representation demands of the League, and the League supporting the Congress demand of Swaraj.

With the end of the war and with the rise of the modernist Kamal Pasha in Turkey, the League and the British, in India, had nothing against each other. So the League began going away from the Congress and began coming closer to the Britishers in India. Thereafter, started the dominance of the fundamentalists in the League, the distance between the League and the Congress kept widening.

Responding to the challenge posed by the Secretary of State, Lord Birkenhead, the All-Party Conference (1928) prepared a Constitution for India which the League supported but which it rejected in the League’s son, there came up Jinnah’s 14 points. These 14-points constituted nothing but separate electorate representation. Their demand for separate electorate representation kept increasing year after year.

As the League’s public support was never strong, it did not obtain much electoral victories in the elections held in 1937 under the Government of India Act, 1935. The league had its government only in two provinces. When the Congress ministries resigned in protest against the Viceroy’s declaration of war in 1939, the League organised the day of deliverance. In 1910, the Muslim League encouraged by the Britishers, got passed the Pakistan resolution at its Lahore session in 1940.

Question 41.
Write notes on the following: Simon Commission; 26 January 1930; Ghadar Party; India League; the Indian National”Army; the Meerut and Lahore Conspiracy Cases.
Answer:
Simon Commission: The commission, known as Simon Commission was appointed in 1927 which visited India in 1928 to enquire into the functioning of the diarchy as proposed by the Government of India Act, 1919. The Swaraj Party also hastened the appointment of Simon Commission.

The Congress boycotted the Simon Commission for it had no faith in it because it was wholly constituted by all-Englishmen. The commission was greeted by all black flags with slogans such as “Simon, go back”. However, the government came up with lathi blows against the demonstrators.

26th January 1930: This day has been an important day in the history of the Indian nationalist movement. It was on this day that the Indians celebrated the unilateral independence day and thereafter every year. In the Lahore Congress Session (1929) the Congress had fixed January 26, 1930 to celebrate the unilateral independence day. But when India became independent on August 15, 1947 the 26th January came to be celebrated as the Republic Day.

The Ghadar Party: The Ghadar, meaning the Rebellion Party was formed in 1913 by the Indian revolutionaries in the United States of America and Canada. Both the Hindus and the Muslims manned this party. The Ghadar party was pledged to wage the war of independence against the British rule in India. The Ghadarites decided to send money and aminurutions to India to help the revolutionaries fight the government. They were always ready to help the Indian revolutionaries.

India League: The Indians outside India, established the India League. The headquarter of the Indian League was at London its chief objectives were to raise the voice of India and her problems in England and create public opinion in India’s favour. Numerous Englishmen had supported the India League.

The Indian National Army (INA): The Indian National Army was reorganised by Netaji Subash Chandra Bose in Singapore in 1943. The army consisted of the Indians settled in southeast Asia and their soldiers who had left the British royal army in India.

Bose made the army strong and powerful and wanted to free India through armed forces. During the war and with the help of Asia (especially the Japanese) forces captured some areas in the North-East India He gave the Indians slogans such as Delhi Chalo’, ‘Jai Hind’, ‘Give me blood, I will give you freedom’. Some INA officials (Shah Nawaz, Sehgal and Dhillon) were tried after the war, but the Government had to bow to the rising and emotional nationalism created by the Hindu Muslim Sikh unity in 1945.

The Meerut and Lahore conspiracy cases: The Meerut conspiracy case was related to the arrest of some labour leaders in -1920s. The Lahore conspiracy case was related to two members of the Hindustan Republican Socialist Association – Bhagat Singh and B.K.Dutt in early 1930. The two cases aroused intense nationalism The Indian leaders provided legal assistance in both the cases. Whereas the Lahore leaders were freed, Bhagat Singh, Sukh Dev and Rajguru were sentenced to death.

Question 42.
“The British were responsible for both the unification and the partition of India.” Explain this statement.
Answer:
The British came to India as traders and became rulers. Following the revolt of 1857, the British Government took over the administration of India. The British rule, in India, was constructive as well as destructive. The British rule, in India, was constructive in the sense that it brought about political unity in the country.

Through uniformity of laws and through the use of English language, the British rulers brought the Indians on a common platform where they came to know the imperialist designs of the British rule in India, Through the English language, the Indians came together and shared their suffering.

But at the same time, the British rule in India was destructive It was destructive in the sense that it divided the communities among themselves. The Britishers exploited the Hindus against the Muslims, and the Muslims against the Hindus. They pursue the policy of divide and rule and encouraged the Muslim League to keep seeking demands after demands. The result was the partition of India.

Question 43.
What led to the launching of the Non-Cooperation Movement in 1920? State as to how non-cooperation movement was effective?
Answer:
The causes to launch the Non-Cooperation Movement in 1920 Gandhiji wanted that the national movement must be led by all Hindu and Muslims. He got an opportunity of making the national movement of Hindus and Muslims. The Muslims were also annoyed by the British Government.

The British government did not give proper treatment to Sultan of Turkey though Indian Muslim helped Britishers in the first World War Indian Muslims decided to agitate. Thus, along with Non-cooperation Movement, Khilafat Movement started.

The Non-cooperation Movement launched in 1920 was to be strictly non-violent. The people were asked to boycott the British goods, to use items made by Indians only, to boycott government jobs, government schools, colleges and courts.

The Congress in its Nagpur session advised the people to observe the following during Non-cooperation Movement:

• To surrender titles and honorary offices.
• To resign from nominated seats in the councils.
• To boycott government darbars and other functions.
• To withdraw gradually, children from the government schools and colleges.
• To boycott orders of the courts.
• To refuse to join British forces in Mesopotamia.
• To boycott election to the councils.
• To boycott foreign goods.

Question 44.
Explain the main features of the Non-cooperation Movement in India’s freedom struggle.
Answer:
Gandhiji launched the Non-cooperation Movement in 1920.
The main features of the Non-cooperation movement can be stated as under:

• To boycott all the official functions of the Government.
• To surrender all titles bestowed by the Government.
• To ask the Indians to boycott British schools, colleges and courts.
• To denounce all the retrogressive measures adopted by the Government.
• To demand ‘Swaraj’ from the Britishers.
• To emphasise on the Hindu-Muslim unity.

The movement went very well. It shook the very foundations of the British rule in India. Following the use of violence in Chauri Chaura (Gorakhpur). Gandhiji withdrew the movement. The”movement was able to create awakening among the Indians.

Question 45.
Discuss the Civil Disobedience Movement (1930-34).
Answer:
The Civil Disobedience Movement was launched by Mahatma Gandhi in 1930 when he violated the Salt Law at Dandi. It was in accordance with the Poorna Swaraj resolution of 1929. As per the resolve of the Civil Disobedience movement, Gandhiji declared. Let every villager fetch or manufacture, contraband salt, sisters should picket liquor shops, opium dens and foreign cloth dealers’ shops. Foreign cloth should be burnt. Hindus should eschew untouchability.

Under the auspices of the movement, the Hartals were organised, foreign cloth dealers’ shops were picketed, opium and liquor dens were boycotted. As the movement progressed, the leaders of the Indian National Congress were arrested.

The Government resorted to oppressive measures and came heavy on the demonstrators. The Indian National Congress boycotted the First Round Table Conference held in 1930. The movement was called off in view of the Gandhi- Irwin pact of 1931. The movement created political awakening among the Indians and brought independence close to them.

Question 46.
Prove by giving examples that the policy of non-violence adopted by Gandhiji was mainly instrumental in achieving Indian independence. (Any four).
Answer:

• Gandhi’s policy of non-violence was supported by his urge for mass participation in the freedom struggle. He made the Congress a mass Congress. The Congress movement became a movement in which peasants, students, women and others came to participate in the liberation struggle.
• Gandhi’s non-violence meant non-cooperation. In all his movements, Gandhiji advocated non-cooperation and boycott of foreign-made goods.
• Gandhi’s non-violence also meant civil- disobedience. He launched civil disobedience movement in 1930 to press the fulfilment of his demands. He never confronted violence with violence
• Gandhi’s non-violence meant maximum inconvenience for oneself so to give the other maximum convenience.

Question 47.
Why did various classes and groups of Indians participate in the civil disobedience movement?
Answer:
The following classes and social groups of Indians participated in the civil disobedience movement:
(i) In the countryside, rich peasant communities like the Patidars of Gujarat and the Jats of Uttar Pradesh were active in the movement. Being producers of commercial crops, they were very hard hit by the trade depression and falling prices.

As their cash income disappeared, they found it impossible to pay the government’s revenue demand and the refusal of the government to reduce the revenue demand led to widespread resentment. These rich peasants became enthusiastic supporters of the Civil Disobedience Movement.

(ii) The Indian merchants and industrialists had made huge profits and become powerful Keen on expanding their business they now reacted against colonial policies that restricted business activities. They wanted protection against imports of foreign goods, and a rupee-sterling foreign exchange ratio that would discourage imports.

To organise business interests, they formed the Indian Industrial and Commercial Congress in 1920 and the Federation of the Indian Chamber of Commerce told Industries (FICCI) in 1927 Led by prominent industrialists like Pursholtamdas Thakurdas and GD, Birla, the industrialists attacked colonial control over the Indian economy and supported the Civil Disobedience Movement when it was first launched. They gave financial assistance and refused to buy or sell imported goods.

(iii) The industrial working classes did not participate in the Civil Disobedience Movement in large numbers, except in the Nagpur region. As the industrialists came closer to the Congres workers stayed aloof. But in spite of that, some workers did participate in the Civil Disobedience Movement, selectively adopting some of the ideas of the Gandhian programme, like boycott of foreign goods, as part of their own movements against low wages and poor working conditions. There were strikes by railway workers in 1930 and dockworkers in 1932. In 1930 thousands of workers in Chotaragpur tin mines wore Gandhi caps and participated in protest rallies and boycott campaigns.

(iv) Another important feature of the Civil Disobedience Movement was the large-scale participation of women. During Gandhiji’s salt march, thousands of women came out of their homes to listen to him. They participated in protest marches, manufactured salt, and picketed foreign cloth and liquor shops. Many went to jail.

In urban areas these women were from high-caste families; in rural areas, they came from rich peasant households. Moved by Gandhiji’s call they began to see service to the nation as a sacred duty of women.

Question 48.
State the main features of the Indian National Movement from 1942 to 1947.
Answer:
The main features of the Indian National Movement from 1942 to 1947 can be stated as under:

• The failure of the Cripps proposal (April 1942) disappointed many Indians. They thought of adopting harsher methods.
• The continuous failures of the Allied Powers and the success of the Axis powers, especially of japan in Asia gave the Indians enough courage to state their demands in clearer terms.
• The Quit India Resolution of August 1942, following the arrest of all the members of the Congress Working Committee, awakened tremendous consciousness among the people.
• The role of Indian National Army (Azad Hind Fauj) led by Subhash Chandra Bose fought for India’s independence from abroad.
• The victory of the Allied powers and of the Labour Party in English elections created hopes among the Indians for an early resolution of India’s problems.
• The INA trials, the Mountbatten plan and the Indian Independence Act all brought the Indians close to their independence.

Question 49.
Why did the Indian National Congress change its goal from Swaraj to Complete Swaraj?
Answer:
The Indian National Congress changed its goal of Swaraj (1906) to Complete Swaraj (1929) because of numerous reasons Some such causes can be summed up as under:

• The pressure of the younger generation of the Indian National Congress was mounting in favour of Complete Swaraj year after year.
• The British Government in India, kept pursuing the policy of divide and rule thus making, the Congress believe that it should launch a – movement for greater demand. Complete Swaraj instead of Swaraj.
• The demand of Complete Swaraj had the backing of the masses while that of Swaraj, the demand of the few English educated people.
• The demand of Complete Swaraj was the only alternative to the communalistic politics of the Muslim League.

Objective Type Questions

1. Fill in the blanks with the appropriate words:

Question 1.
The national song, ……………………… was composed by Bankim Chandra Chatterjee.
Answer:
Vande Matram.

Question 2.
……………………… was the architect of two-nation theory.
Answer:
Jinnah.

Question 3.
The Gandhi-Irwin pact was signed in
Answer:
1931.

2. Match the following:

A	B
Manford Reforms	C.R. Das
Swarajya party	Khilafat Movement
All Brothers	Subhash Chandra Bose
Gandhi Ji	1919
Azad Hind Fauj	Dyer
Jalliawala Massacre	Civil Disobedience Movement

Answer:

A	B
Manford reforms	1919
Swarajya party	C.R. Das
Ali Brothers	Khilafat Movement
Gandhiji	Subhash Chandra Bose
Azad Hind Fauj	Civil Disobedience Movement
Jalliawala Massacre	Dyer

Question 3.
Choose the most appropriate alternative:
(i) The Rowlatt Act was proposed in:
(a) 1917
(b) 1918
(c) 1919
(d) 1920
Answer:
(c) 1919

(ii) One of the following was not the earlier satyagraha launched by Gandhiji:
(a) Individual
(b) Champaran
(c) Kheda
(d) Ahmedabad
Answer:
(a) Individual

(iii) Dr Iqbal presided the Muslim League session in:
(a) 1930
(b) 1931
(c) 1932
(d) 1933
Answer:
(a) 1930

(iv) A pact resolving the communal issue was undertaken in 1932. Where was the pact signed?
(a) Patna
(b) Poona
(c) Peshwar
(d) Patti
Answer:
(b) Poona

(v) Anandmath was written by:
(a) Rabindranath Tagore
(b) Bankim Chandra Chattopadhyay
(c) Abanindranath Tagore
(d) Debandranath Tagore
Answer:
(b) Bankim Chandra Chattopadhyay.

Extra Questions for Class 10 Social Science

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Class 10 Science Chapter 1 Extra Questions and Answers Chemical Reactions and Equations

Extra Questions for Class 10 Science Chapter 1 Chemical Reactions and Equations with Answers Solutions

Extra Questions for Class 10 Science Chapter 1 Very Short Answer Type

Class 10 Science Chapter 1 Extra Questions Question 1.
How does the food become rancid?
Answer:
Food becomes rancid when fat and oils present in the food are oxidised.

Chemical Reaction And Equation Class 10 Extra Questions Question 2.
A student burnt a metal A found in the form of ribbon. The ribbon burnt with a dazzling flame and a white powder B was formed which was basic in nature. Identify A and B. Write the balanced chemical equation.
Answer:
X = Mg,  Y = MgO,  Mg + O₂ → 2MgO

Chemical Reactions And Equations Extra Questions Question 3.
What is a balanced chemical equation?
Answer:
An equation that has equal number of atoms of each element on both the sides of the equation is called a balanced chemical equation, i.e., mass of the reactants is equal to mass of the products.
For example, \(2 \mathrm{Mg}+\mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{MgO}\)

Chemical Reactions And Equations Class 10 Extra Questions Answers Question 4.
Write a balanced equation for a chemical reaction that can be characterised as precipitation.
Answer:
BaCl₂(aq) + Na₂SO₄ (aq) → BaSO₄(s) + 2NaCl(aq)

Chemical Reactions And Equations Class 10 Extra Questions Pdf Question 5.
What is rust?
Answer:
It is a brown mass known as hydrated ferric oxide. Its formula is Fe₂O₃. xH₂O.

Extra Questions Of Chemical Reactions And Equations Question 6.
A zinc rod is left for nearly 20 minutes in a copper sulphate solution. What change would you observe in the zinc rod?
Answer:
The zinc rod will change into zinc sulphate.

Chemical Reaction And Equation Extra Questions Question 7.
Name two salts that are used in black and white photography.
Answer:
Both silver chloride and silver bromide are used in black and white photography.

Extra Questions For Class 10 Science Chapter 1 Question 8.
Which chemical process is used for obtaining a metal from its oxide?
Answer:
The process is known as the reduction of metal oxide.

Class 10 Chemical Reactions And Equations Extra Questions Question 9.
If you collect silver coins and copper coins you may have seen that after some days a black coating forms on silver coins and a green coating on copper coins. Which chemical phenomenon is responsible for these coatings? Write the chemical name of the black and green coatings.
Answer:
Corrosion is responsible for the formation of this coating. Black coating is due to formation of Ag₂S and green coating is due to formation of CuCO₃.Cu(OH)₂.

Class 10 Chemistry Chapter 1 Extra Questions Question 10.
When carbon dioxide is passed through lime water, it turns milky, why?
Answer:
Lime water (calcium hydroxide) combines with carbon dioxide to form a suspension of calcium carbonate which makes lime water milky.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Chemical Reactions And Equations Class 10 Extra Questions Question 11.
Identify the most reactive and least reactive metal: Al, K, Ca, Au.
Answer:
Most reactive metal: K(Potassium); least reactive metal: Au(gold).

Class 10 Science Ch 1 Extra Questions Question 12.
X + Y SO₄ → X SO₄ + Y
Y + X SO₄ → No reaction
Of the two elements T and Y which is more reactive and why?
Answer:
‘X’ is more reactive than ‘Y since it has displaced ‘Y’ in the displacement reaction.

Chapter 1 Science Class 10 Extra Questions Question 13.
Why is it necessary to balance a chemical equation?
Answer:
An equation is balanced in order to satisfy the law of conservation of mass according to which total mass of the reactants is equal to the total mass of the products, i.e., mass can neither be created nor be destroyed during any chemical change.

Class 10 Science Chapter 1 Extra Questions With Solutions Question 14.
During electrolysis of water, the gas collected in one test tube is double than the other, why?
Answer:
On electrolysis, water decomposes into hydrogen and oxygen in the ratio 2 : 1 by volume so, H₂ gas collected in one test tube is double than O₂.

Extra Questions For Class 10 Science Chapter 1 With Answers Question 15.
Represent decomposition of ferrous sulphate with the help of balanced chemical equation.
Answer:
2FeSO₄ (s) → Fe₂O₃ (s) + SO₂ (g) + SO₃ (g)

Question 16.
What is a chemical equation?
Answer:
A chemical equation is a symbolic notation that uses formulae instead of words to represent a chemical equation.

Question 17.
A teacher took a few crystals of sugar in a dry test tube and heated the test tube over a flame. The colour of sugar turned black. Explain why?
Answer:
Sugar is a complex compound which on heating undergoes decomposition. Water gets evaporated thereby leaving behind black carbon in the test tube.

Question 18.
Name two metals which do not get corroded.
Answer:
Gold (Au) and platinum (Pt) do not get corroded.

Question 19.
Identify the compound oxidised in the following reaction:
H₂S (g) + Cl₂ → S(s) + 2HCl (g)
Answer:
H₂S is oxidised.

Question 20.
Why is a magnesium ribbon cleaned before burning?
Answer:
Magnesium reacts with moist air and forms a layer of oxide, MgO (white), on its surface. So, a magnesium ribbon is cleaned to remove the oxide layer before burning.

Question 21.
State the chemical change that takes place when limestone is heated.
Answer:
Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide.

Question 22.
On which chemical law, balancing of chemical equation is based?
Answer:
Balancing of a chemical equation is based on the law of conservation of mass.

Question 23.
Name the term used for the solution of the reactants or products when dissolved in water.
Answer:
Aqueous

Question 24.
What happens when magnesium ribbon burns in air?
Ans. When magnesium ribbon burns in air, it combines with the oxygen to form magnesium oxide.
2Mg(s) + O₂(g) → 2MgO(s)

Question 25.
Give an example of an exothermic reaction.
Answer:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) + Heat (evolved)

Question 26.
What type of reaction is this: Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl
Answer:
It is a double displacement reaction.

Question 27.
Give an example of exothermic reaction.
Answer:
N₂ (g) + O₂ (g) → 2NO (g) – Heat (absorbed)

Question 28.
A substance X used for coating iron articles is added to a blue solution of a reddish brown metal Y. The colour of the solution gets discharged. Identify X and Y and also the type of reaction.
Answer:
X = Zn, Y = Cu, Displacement reaction.

Question 29.
Name the gas evolved when zinc reacts with dil. HCl.
Answer:
Hydrogen gas is evolved.

Extra Questions for Class 10 Science Chapter 1 Short Answer Type I

Question 1.
You are given the following materials
(i) Marble chips (ii) dilute hydrochloric acid (iii) Zinc granules
Identify the type of reaction when marble chips and zinc granules are added separately to acid taken in two test tubes.
Answer:
(i) Marble chips react with dilute hydrochloric acid to form calcium chloride and carbon dioxide. It is a double displacement reaction.
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

(ii) Zinc granules react with dilute hydrochloric acid to give hydrogen gas. It is a displacement reaction.
Zn (s) + 2HCl → ZnCl₂ (aq) + H₂ (g)

Question 2.
What do you understand by precipitation reaction? Explain with suitable examples.
Answer:
The reaction in which two compounds in their aqueous state react to form an insoluble compound. When two reactants react and product formed remains insoluble and settles as a solid it is substance (precipitate) is called a precipitation reaction.

For example,
(i) When aqueous solution of sodium sulphate is mixed with an aqueous solution or barium chloride, barium sulphate is obtained as a white precipitate.
Na₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) + 2NaCl (ag)

(ii) When aqueous solution of sodium chloride is mixed with an aqueous solution of silver nitrate, silver chloride is obtained as a white precipitate.

Question 3.
What happens when aqueous solutions of sodium sulphate and barium chloride are mixed? What type of reaction is it?
Answer:
On mixing the solutions of sodium sulphate and barium chloride, a white precipitate of barium sulphate is obtained.

It is a double displacement reaction.

Question 4.
Explain the following terms with suitable examples.
(a) Oxidation
(b) Reduction
Answer:
(a) Oxidation is a process of addition of oxygen to a substance or removal of hydrogen from a substance, for example,

Copper is oxidised to CuO, as oxygen is added to copper.

(b) It is the process of removal of oxygen from a substance or addition of hydrogen to a substance, for example,

Copper oxide is reduced to copper as it involves removal of oxygen.

Question 5.
Complete the missing components/variables given as x and y in the following reactions.  [NCERT Exemplar]
(a) Pb(NO₃)₂ (aq) + 2Kl (aq) → PbI₂ (x) + 2KNO₃ (y)
(b) Cu (s) + 2AgNO₃ (aq) → Cu(NO₃)₂ (aq) + x (s)
(c) Zn (s) + H₂SO₄ (aq) → ZnSO₄ (x) + H₂ (y)

Answer:
(a) x = (s), y = (aq)
(b) x = 2Ag
(c) x = (aq); y = (g)
(d) x = heat

Question 6.
An iron knife kept dipped in a blue copper sulphate solution turns the blue solution light green. Why?
Answer:
We know that iron is more reactive than copper, so it displaces copper from copper sulphate solution and forms ferrous sulphate which is of light green colour.

Question 7.
A, B and C are three elements which undergo chemical reactions in the following way.
A₂O₃ + 2B → B₂O₃ + 2A
3CSO₄ + 2B → B₂(SO₄)₃ + 3C
3CO + 2A → A₂O₃ + 3C
Answer the following:
(a) Which element is most reactive?
(b) Which element is least reactive?
Answer:
(a) The most reactive element is ‘B’. It has displaced both ‘A’ and ‘C’ from their compounds.
(b) The least reactive element is ‘C’ as it has been displaced by both ‘A’ and ‘B’.

Question 8.
Write the balanced chemical equations for the following reactions and identify the type of reaction in each case.
(a) Nitrogen gas is treated with hydrogen gas in the presence of a catalyst at 773 K to form ammonia gas.
(b) Sodium hydroxide solution is treated with acetic acid to form sodium acetate and water.
(c) Ethanol is warmed with ethanoic acid to form ethyl acetate in the presence of concentrated H₂SO₄.
(d) Ethene is burnt in the presence of oxygen to form carbon dioxide, water and releases heat and light.  [NCERT Exemplar]
Answer:

Question 9.
What is lime water test for the detection of carbon dioxide?
Answer:
When carbon dioxide gas is passed through lime water, it turns milky due to the formation of a milky suspension (precipitate) of calcium carbonate.
Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) + CO₂ (g)

Question 10.
During the reaction of some metals with dilute hydrochloric acid, following observations were made,
(a) Silver metal does not show any change.
(b) The temperature of the reaction mixture rises when aluminium (Al) is added.
(c) The reaction of sodium metal is found to be highly explosive.
(d) Some bubbles of a gas are seen when lead (Pb) is reacted with the acid.
Explain these observations giving suitable reasons.   [NCERT Exemplar]
Answer:
(a) No change takes place because silver metal does not react with hydrochloric acid in normal situations.
(b) The reaction between hydrochloric acid and aluminium is exothermic, thus the temperature of the reaction mixture rises when aluminium is added.
(c) Since, sodium is a highly reactive metal, thus it reacts with hydrochloric acid vigorously and produces a large amount of heat. Thus, the reaction is exothermic.
(d) Bubbles of hydrogen gas are formed when lead react with dilute hydrochloric acid.
Pb + 2HCl → PbCl₂ + H₂

Question 11.
A copper coin is kept in a solution of silver nitrate for some time. What will happen to the coin and the colour of the solution?
Answer:
We know that copper is more reactive than silver, so it will displace silver from its salt solution.
Cu (s) + 2AgNO₃ (aq) → Cu(NO₃)₂ (aq) + 2Ag (s)
So, the solution will turn blue due to the formation of copper nitrate.

Question 12.
An aqueous solution of metal nitrate P reacts with sodium bromide solution to form yellow ppt. of compound Q which is used in photography. Q on exposure to sunlight undergoes decomposition reaction to form metal present in P along with reddish brown gas. Identify P and Q. Write the chemical reaction and type of chemical reaction.
Answer:
P = Ag NO₃, Q = AgBr
AgNO₃ (aq) + NaBr (aq) → NaNO₃ (aq) + AgBr (s); Double decomposition reaction
2AgBr (s) → 2Ag (s) + Br₂ (g); Photochemical decomposition reaction

Question 13.
What happens when iron nails are immersed in copper sulphate solution? What type of reaction is it?
Answer:
When iron nails are immersed in copper sulphate solution, iron ions displace copper ions and a new compound ferrous sulphate is formed.

It is a diplacement reaction.

Question 14.
Which of the following reaction is possible. Explain giving suitable reason.
(i) Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)
(ii) Fe (s) + ZnSO₄ (aq) → FeSO₄ (aq) + Zn (s)
(iii) Zn (s) + FeSO₄ (s) → ZnSO₄ (aq) + Fe (s)
Answer:
Reaction (i) and (iii) are possible.
(i)

Zinc is more reactive than copper, therefore, it can displace copper from copper sulphate solution.

(iii) Zn (s) + FeSO₄ (aq) → ZnSO₄ (aq) + Fe (s)
Zn is more reactive than Fe, therefore, it can displace iron from ferrous sulphate solution.

Reaction (ii) is not possible as iron is less reactive than zinc, hence, it cannot displace Zn.

Question 15.
Which among the following changes are exothermic or endothermic in nature?
(a) Decomposition of ferrous sulphate
(b) Dilution of sulphuric acid
(c) Dissolution of sodium hydroxide in water
(d) Dissolution of ammonium chloride in water  [NCERT Exemplar]
Answer:
(a) endothermic
(b) exothermic
(c) exothermic
(d) endothermic.

Question 16.
State one advantage and one disadvantage of corrosion.
Answer:
Advantage. In some metals a protective layer is formed on its surface due to corrosion which prevent it from further corrosion.

Example, Aluminium (Al) forms a layer of aluminium oxide (Al₂O₃) by corrosion. This layer prevents further corrosion of aluminium.
Disadvantage: Loss of the metal.

Extra Questions for Class 10 Science Chapter 1 Short Answer Type II

Question 1.
What is corrosion? State condiyions necessary for rusting of iron. How is rusting harmful?  [NCERT Exemplar]
Answer:
Corrosion. The process of eating away of the metal surface by the action of atmospheric reagents like water, oxygen and acids changing the metal into its compound is called corrosion.

Rusting of iron. When iron objects are exposed to atmosphere, they are attacked by air and moisture (water) of the atmosphere and a brown and orange coloured layer is formed on the surface. It is called rust which is mainly hydrated iron (III) oxide Fe₂O₃. xH₂O.

Harmful effect of rusting. Hydrated iron (III) oxide is a brittle substance and falls off from the surface of iron and thus the object is damaged. Holes, cavities and roughness of surfaces are the result of rusting of an iron object.

Conditions necessary for rusting:

• Open surface of the metal
• Presence of air (oxygen)
• Presence of moisture (water).

Question 2.
The gases hydrogen and chlorine do not react with each other even if kept together for a long time. However, in the presence of sunlight, they readily combine. What actually happens?   [NCERT Exemplar]
Answer:
In chemical reactions, energy is needed to break the bonds present in the reacting molecules so that they may combine to form the products. In this reaction, sunlight is the source of energy in the form of photons. The energy made available by sunlight helps in breaking the bonds and this leads to chemical reaction between hydrogen and chlorine.

Question 3.
Write the balanced chemical equations for the following reactions and identify the type of reaction in each case.
(a) Thermite reaction, iron (III) oxide reacts with aluminium and gives molten iron and aluminium oxide.
(b) Magnesium ribbon is burnt in an atmosphere of nitrogen gas to form solid magnesium nitride.
(c) Chlorine gas is passed in an aqueous potassium iodide solution to form potassium chloride solution and solid iodine.
(d) Ethanol is burnt in air to form carbon dioxide, water and releases heat.   [NCERT Exemplar]
Answer:
The balanced equations are as under:
(a) Fe₂O₃ (s) + 2Al (s) → 2Fe (l) + Al₂O₃ (s) + Heat
It is a redox reaction / displacement reaction.

(b) 3Mg (s) + N₂ (g) → Mg₃N₂ (s)
It is a combination reaction as well as redox reaction.

(c) Cl₂ (g) + 2KI (aq) → 2KCl (aq) + I₂ (s)
It is a displacement reaction as well as redox reaction.

(d) C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (l) + Heat
Redox reaction/combustion reaction

Question 4.
What is rancidity? Write the common method to prevent it.
Answer:
When food item are kept unprotected for some time, their smell and taste changes. This process is called rancidity. Actually, the microorganisms oxidise the fat and oils present in them. So oxidation of food items need to be prevented to protect them.

Common methods to prevent rancidity of food item:

• Keeping the food at low temperature.
• Keeping food item in air tight containers.
• By filling nitrogen in the food storage bags.

Question 5.
(a) What happens chemically when quick lime is added to water?
(b) Write the chemical equation in balanced form.
MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O
(c) What is decomposition reaction? Explain it with a suitable example.
Answer:
(a) When quick lime (CaO) is added to water, slaked lime Ca(OH)₂ is formed. The reaction is highly exothermic in nature.

(b) The balanced chemical equation is:
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

(c) Decomposition reaction is a chemical reaction in which a single substance splits or breaks into two or more substances under suitable conditions. For example,
2FeSO₄(s) → Fe₂O₃(s) + SO₂(g) + SO₃ (g)

Question 6.
Identify the oxidising agent (oxidant) in the following reactions:
(a) Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O
(b) 2Mg + O₂ → 2MgO
(c) CuSO₄ + Zn → Cu + ZnSO₄
(d) V₂O₅ + 5Ca → 2V + 5CaO
(e) 3Fe + 4H₂O → Fe₃O₄ + 4H₂
(f) CuO + H₂ → Cu + H₂O [NCERT Exemplar]
Answer:
(a) Pb₃O₄
(b) O₂
(c) CuSO₄
(d) V₂O₅
(e) 4H₂O
(f) CuO

Question 7.
Write the balanced chemical equations for the following reactions:  [NCERT Exemplar]
(a) Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogencarbonate.
(b) Sodium hydrogencarbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide.
(c) Copper sulphate on treatment with potassium iodide precipitates cuprous iodide (Cu₂I₂), liberates iodine gas and also forms potassium sulphate.
Answer:
(a) Na₂CO₃ + HCl → NaCl + NaHCO₃
(b) NaHCO₃ + HCl → NaCl + H₂O + CO₂
(c) 2CUSO₄ + 4KI → 2K₂SO₄ + Cu₂I₂ + I₂

Question 8.
Write chemical equations for the reactions taking place when:
(i) Iron reacts with steam
(ii) Magnesium reacts with dilute HCl
(iii) Copper is heated in air
Answer:

Question 9.
fialance the following chemical equations and identify the type of chemical reaction.
(a) Mg (s) + Cl₂ (g) → MgCl₂ (s)

(d) TiCl₄ (l) + Mg (s) → Ti (s) + MgCl₂ (s)
(e) CaO (s) + SiO₂ (s) → CaSiO₃ (s)

[NCERT Exemplar]
Answer:
The chemical equation in their balanced form may be written as follows:
(a) Mg (s) + Cl₂ (g) → MgCl₂ (s), Combination reaction

(d) TiCl₄ (l) + 2Mg (s) → Ti (s) + 2MgCl₂ (s); Displacement reaction
(e) CaO (s) + SiO₂ (s) → CaSiO₃ (s); Combination reaction

Question 10.
A silver article generally turns black when kept in the open for a few days. The article when rubbed with toothpaste again starts shining.
(a) Why do silver articles turn black when kept in the open for a few days? Name the phenomenon involved.
(b) Name the black substance formed and give its chemical formula.   [NCERT Exemplar]
Answer:
(a) Silver articles turn black when kept in the air for a few days because H₂S gas present in the air attacks silver forming a coating of black silver sulphide. The phenomenon is called corrosion.

(b) Black substance formed is silver sulphide (Ag₂S)
2Ag (s) + H₂S (g) → Ag₂S (s) + H₂ (g).

Question 11.
Which among the following are physical or chemical changes?
(a) Evaporation of petrol
(b) Burning of Liquefied Petroleum Gas (LPG)
(c) Heating of an iron rod to red hot
(d) Curdling of milk
(e) Sublimation of solid ammonium chloride  [NCERT Exemplar]
Answer:
(a) Physical change
(b) Chemical change
(c) Physical change
(d) Chemical change
(e) Physical change

Extra Questions for Class 10 Science Chapter 1 Long Answer Type

Question 1.
Balance the following equations:
(a) Bacl₂ + H₂SO₄ → BaSO₄ + HCl
(b) CH₄ + O₂ → CO₂ + H₂O



Answer:
(a) Bacl₂ + H₂SO₄ → BaSO₄ + 2HCl
(b) CH₄ + 2O₂ → CO₂ + 2H₂O

Question 2.
On heating blue coloured powder of copper (II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed.
(а) Write a balanced chemical equation of the reaction.
(b) Identify the brown gas X evolved.
(c) Identify the type of reaction.
(d) What could be the pH range of aqueous solution of the gas X?  [NCERT Exemplar]
Answer:

(b) Brown gas X is nitrogen dioxide (NO₂).
(c) It is a thermal decomposition reaction.
(d) The gas (NO₂) is an oxide of a non-metal. Hence, its aqueous solution will be acidic, i.e., pH range would be between 0 and 7.

Question 3.
(A) Name the type of chemical reaction represented by the following equation:

(c) Zn(s) + H₂SO₄(ag) → ZnSO₄(aq) + H₂(g)
(B) “A solution of potassium chloride when mixed with silver nitrate solution, and an insoluble white substance is formed”.   [CBSE 2010, 2012]
(i) Translate the above statement into a chemical equation.
(ii) State two types for the classification of this reaction.
Answer:
(A) (a) Decomposition reaction
(b) Combination reaction
(c) Displacement reaction.

(B) (i) KCl (aq) + AgNO₃ (aq) → AgCl (s) + KNO₃ (aq)
(ii) It is a double displacement reaction also called precipitation reaction.

Question 4.
What happens when zinc granules are treated with dilute solution of H₂SO₄, HCl, HNO₃, NaCl and NaOH, also write the chemical equations if reaction occurs.  [NCERT Exemplar]
Answer:

Question 5.
(A) Write the balanced chemical equations for the following chemical reactions:
(i) Hydrogen + Chlorine → Hydrogen chloride
(ii) Lead + Copper chloride → Lead chloride + Copper
(iii) Zinc oxide + Carbon → Zinc + Carbon monoxide
(B) Write balanced chemical equations for the following reactions:
(a) Silver bromide on exposure to sunlight decomposes into silver and bromine.
(b) Sodium metal reacts with water to form sodium hydroxide and hydrogen gas.
Answer:
(A) (i) H₂ + Cl₂ → 2HCl
(ii) Pb + CuCl₂ → PbCl₂ + Cu
(iii) ZnO + C → Zn + CO
(B)

(b) 2Na + 2H₂O→ 2NaOH + H₂

Question 6.
(a) Why cannot a chemical change be normally reversed?
(b) Why is it always essential to balance a chemical equation?
(c) What happens when CO₂ gas is passed through lime water and why does it disappear on passing excess CO₂?
(d) Can rusting of iron takes place in distilled water?
Answer:
(a) In a chemical change some bonds are broken and some bonds are formed. The products are quite different from the reactants. Therefore, it normally can’t be reversed.
(b) A chemical equation has to be balanced to satisfy the law of conservation of mass.
(c) On passing CO₂ gas through lime water, it turns milky due to formation of insoluble calcium carbonate which dissolves on passing excess CO₂ due to formation of soluble calcium bicarbonate.
Ca(OH)₂ + CO₂(g) → CaCO₃(s) + H₂O(l)
CaCO₃(s) + H₂O(l) + CO₂(g) → Ca(HCO₃)₂ (soluble)
(d) No

Question 7.
What happens when a piece of
(a) zinc metal is added to copper sulphate solution?
(b) aluminium metal is added to dilute hydrochloric acid?
(c) silver metal is added to copper sulphate solution?
Also, write the balanced chemical equation if the reaction occurs.  [NCERT Exemplar]
Answer:
(a) Zinc is more reactive than copper. It displaces Cu from CuSO₄ solution forming colourless zinc sulphate. Thus, blue colour of CuSO₄ solution starts fading and ultimately blue colour disappears.

(b) Aluminium reacts with dilute HCl acid forming AlCl₃ along with evolution of bubbles of H₂ gas.

(c) Silver is less reactive than copper. Hence, Ag cannot displace Cu from CuSO₄ solution. Thus, no reaction occurs.

Question 8.
(A) A brown substance ‘X’ on heating in air forms a substance ‘Y’. When hydrogen gas is passed over heated ‘Y’, it again changes back into ‘X’.
(i) Name the substances X and Y.
(ii) Name the chemical processes occurring during both the changes.
(iii) Write the chemical equations.   [CBSE 2011]
(B) A metal is treated with dil. H₂SO₄. The gas evolved is collected by the method shown in the figure. Answer the following:

(i) Name the gas.
(ii) Name the method of collection of the gas.
(iii) Is the gas soluble or insoluble in water?
(iv) Is the gas lighter or heavier than air?
Answer:
(A) (i) The substance X is copper and Y is copper (II) oxide or CuO.
(ii) The process of change of X into Y is oxidation. The process of change of Y into X is reduction.
(iii) The chemical equations are:

(B) (i) The gas evolved is hydrogen.
(ii) The method of collection of the gas is the downward displacement of water.
(iii) The gas is insoluble in water. That is why, it can be collected over water.
(iv) The gas is lighter than air.

Chemical Reactions and Equations HOTS Questions With Answers

Question 1.
The marble statues often slowly get corroded when kept in open for a long time. Assign a suitable explanation.
Answer:
SO₂, NO₂ gases are released into the atmosphere from various sources. These dissolve in rain water to give acid which corrodes marble statues.
2SO₂ + O₂ → 2SO₃
H₂O + SO₃ → H₂SO₄
2NO₂ + H₂O → 2HNO₃
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂

Question 2.
Observe the following activity and identify the following:

(a) Type of chemical reaction.
(b) Write a chemical equation to represent the above change.
(c) Name the gas evolved in the above reaction.
(d) Do you observe any colour change in the above reaction? If yes, mention the colour change.
(e) Name the coloured substance formed.
Answer:
(a) Decomposition reaction.
(b) 2Pb(NO₃)₂ (s) → 2PbO (s) + 4NO₂ (g) + O₂ (g)
(c) Reddish brown fumes of nitrogen dioxide along with oxygen gas is evolved.
(d) The colour of the substance changes from white to yellow due to formation of lead oxide.
(e) Lead oxide which is yellow coloured.

Question 3.
A substance X, which is an oxide of a group 2 element, is used intensively in the cement industry. This element is present in bones also. On treatment with water it forms a solution which turns red litmus blue. Identify X and also write the chemical reactions involved.   [NCERT Exemplar]
Answer:
The above information suggests that the substance ‘A’ is oxide of the element calcium (Ca) which is present in group 2 of the periodic table. Calcium is also a constituent of our bones in the form of calcium phosphate. Calcium oxide (CaO) reacts with water to form calcium hydroxide (basic in nature). It forms a basic solution which turns red litmus blue.
CaO (s) + H₂O (aq) → Ca(OH)₂ (aq)

Question 4.
On adding a drop of barium chloride solution to an aqueous solution of sodium sulphite, a white precipitate is obtained.
(a) Write a balanced chemical equation of the reaction involved.
(b) What other name can be given to this precipitation reaction?
(c) On adding dilute hydrochloric acid to the reaction mixture, white precipitate disappears. Why? [NCERT Exemplar]
Answer:
(a) A white precipitate of barium sulphite is formed when barium chloride is added to the solution of sodium sulphite.
BaCl₂ (aq) + Na₂SO₃ (aq) → BaSO₃ (s) + 2NaCl (aq)

(b) This precipitation reaction is also a double displacement reaction.

(c) Barium chloride, sulphur dioxide and water are formed when dilute hydrochloric acid is added to this solution of barium sulphate and sodium chloride. Since barium chloride is a soluble substance, thus white precipitate of barium sulphite disappears.
BaSO₃ (s) + HCl (ag) → BaCl₂ (ag) + SO₂ (g) + H₂O

Question 5.
A water insoluble substance ‘X’ on reacting with dilute H₂SO₄ released a colourless and odourless gas accompanied by brisk effervescence. When the gas was passed through water, the solution obtained turned blue litmus red. On bubbling the gas through lime water, it initially became milky and the milkiness disappeared when the gas was passed in excess. Identify the substance ‘X’ and write the chemical equations of the reaction involved.
Answer:
The water insoluble substance ‘X’ is most probably the metal carbonate (CaCO₃). The chemical reaction that were involved are given below.
CaCO₃ (s) + H₂SO₄ (aq) → CaSO₄ (ag) + H₂O (aq) + CO₂ (g)
Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) (milky) + H₂O (l)
CaCO₃ (s) + CO₂ (g) + H₂O (aq) → Ca (HCO₃)₂ (milkiness)

Question 6.
The given set up shows the electrolysis of water taking place.

(a) Identify the gases evolved at cathode and anode.
(b) Why is the amount of gas collected in one of the test tubes double the amount in the other? Name this gas.
(c) How will you test the evolved gases?
Answer:
(a) The gases evolved at anode and cathode are oxygen and hydrogen.

(b) On electrolysis, water decomposes to form hydrogen and oxygen gas in the ratio of 2 : 1 by volume according to the equation given below:

The gas collected in double volume is hydrogen.

(c) Test for oxygen. If we bring a burning splinter near the mouth of test tube containing oxygen gas it burns more brightly.
Test for hydrogen. On bringing a burning splinter near the mouth of the test tube containing hydrogen, the gas burns with a pop sound.

Question 7.
A magnesium ribbon is burnt in oxygen to give a white compound X accompanied by emission of light. If the burning ribbon is now placed in an atmosphere of nitrogen, it continues to burn and forms a compound Y.
(a) Write the chemical formulae of X and Y.
(b) Write a balanced chemical equation, when X is dissolved in water.   [NCERT Exemplar]
Answer:

Question 8.
You are provided with two containers made up of copper and aluminium. You are also provided with solutions of dilute HCl, dilute HNO₃, ZnCl₂ and H₂O. In which of the above containers these solutions can be kept? [NCERT Exemplar]
Answer:
(A) When solutions are kept in copper container
(a) Dilute HCl
Copper does not react with dilute HCl. Therefore, it can be kept.

(b) Dilute HNO₃
Nitric acid acts as a strong oxidising agent and reacts with copper vessel, therefore cannot be kept.

(c) ZnCl₂
Zinc is more reactive than copper (Cu) therefore, no displacement reaction occurs and hence can be kept.

(d) H₂O
Copper does not react with water. Therefore, can be kept.

(B) When solutions are kept in aluminium containers
(a) Dilute HCl
Aluminium reacts with dilute HCl to form its salt and hydrogen is evolved. Therefore, cannot be kept.
2 Al + 6HCl → 2AlCl₃ + 3H₂

(b) Dilute HNO₃
Aluminium gets oxidised by dilute HNO₃ to form a layer of Al₂O₃ and can be kept.

(c) ZnCl₂
Aluminium being more reactive than zinc can displace zinc ion from the solution. Therefore, the solution cannot be kept.
2Al + 3ZnCl₂ → 2AlCl₃ + 3Zn

(d) H₂O
Aluminium does not react with cold or hot water. Therefore, water can be kept.
Aluminium is attacked by steam to form aluminium oxide and hydrogen.
2Al(s) + 3H₂O(g) → Al₂O₃(S) + 3H₂ (g)

Extra Questions for Class 10 Science Chapter 1 Value Based

Question 1.
Rakesh wanted to give a coating of white wash on the walls of his house. He purchased quick lime (CaO) and dropped it in container of water and immediately applied the same on the walls. In this process, he spoiled his hands and even suffered minor bums. His friend Kapil advised him to keep the container overnight before applying a coating on the wall.
(i) What mistake was committed by the Rakesh?
(iii) Why did he suffer from minor bums?
(iii) How was Kapil’s advice useful?
(iv) What values are exhibited by Kapil?
Answer:
(i) Rakesh should have waited for a few hours because when quick lime dissolves in water, slaked lime is formed and the process is highly exothermic (a lot of energy is released).

(ii) Because the reaction is exothermic and a lot of energy is released and container becomes hot so he suffered from minor burns.

(iii) Quick lime (CaO) reacts with water to form Ca(OH)₂ which is known as slaked lime. The dissolution process is highly exothermic. So a large amount of heat is evolved. By keeping container overnight, the chemical reaction subsides and the heat dissipates. So the coating of slaked lime can be applied safely on the walls.

(iv) Knowledge of Chemistry, helpful and caring nature.

Question 2.
Palak is an eleven year old girl. She purchased a packet of potato chips and had a few from it more than a month back. She had kept away the open packet containing the remaining potato chips then. She wanted to eat the remaining potato chips now. Her elder sister Anjali found that the potato chips were giving out an unpleasant odour. When she put one of them in her mouth, she found that it had an unpleasant taste too. Anjali threw away the packet and did not allow Palak to eat the potato chips.
(a) What name is given to the condition in which potato chips kept in the open for a long time give out unpleasant smell and taste?
(b) Which chemical reaction is responsible for the spoilage of potato chips?
(c) Explain the reason for the unpleasant smell as well as unpleasant taste of the potato chips.
(d) Mention the values exhibited by Anjali.
Answer:
(a) Rancidity.
(b) Oxidation reaction.
(c) Potato chips contain oil. The oil present in potato chips (which have been kept exposed to air for a considerable time) gets oxidised by the oxygen of air. The oxidation products have unpleasant smell and taste. The potato chips are then said to have turned rancid. They become unfit to eat.
(d) (i) Awareness (or knowledge of Chemistry)
(ii) Concern for the health of her sister.

Here we are providing Probability Class 10 Extra Questions Maths Chapter 15 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Probability with Answers Solutions

Extra Questions for Class 10 Maths Chapter 15 Probability with Solutions Answers

Probability Class 10 Extra Questions Very Short Answer Type

Probability Class 10 Extra Questions Question 1.
State true or false and give the reason. If I toss a coin 3 times and get head each tir ne, then I should expect a tail to have a higher chance in the 4th toss.
Solution:
False, because the outcomes ‘head’ and ‘tail are equally likely. So, every time the probability of getting head or tail is \(\frac{1}{2}\)

Class 10 Probability Extra Questions Question 2.
A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since, this situation has only two possible outcomes, so the probability of each is \(\frac{1}{2}\) Justify.
Solution:
True, because the outcomes odd number’ and `even number’ are equally likely here.

Probability Class 10 Extra Questions With Answers Pdf Question 3.
In a family, having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is \(\frac{1}{4}\). Is this correct? Justify your answer.
Solution:
False, because the outcomes are not equally, likely. For no girl, outcome is bbb, for one girl, it is
bgb, gbb, bbg, for two girls, it is bgg, ggb, gbg and for all girls, it is ggg.

Probability Extra Questions Class 10 Question 4.
A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. 15.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reason.
Solution:

False, because the outcome 3 is more likely than the other numbers.

Extra Questions Of Probability Class 10 Question 5.
Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Solution:
Possible outcomes are {HH, HT, TH, TT}.
(exactly one head) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

Extra Questions On Probability Class 10 Question 6.
From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.
Solution:
Number of black queens in a pack of cards = 2
∴ P (black queen) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Probability Extra Questions Question 7.
If P (E) = 0.05, what is the probability of ‘not E’ ?
Solution:
As we know that,
P (E) + P (not E) = 1
P (not E) = 1 – P (E) = 1 – 0.05 = 0.95

Probability Class 10 Extra Questions With Solutions Question 8.
What is the probability of getting no head when two coins are tossed simultaneously?
Solution:
Favourable outcome is TT;
∴ P (no head) = \(\frac{1}{4}\)

Class 10 Maths Probability Extra Questions Question 9.
In a single throw of a pair of dice, what is the probability of getting the sum a perfect square?
Solution:
Total outcomes = 36
Favourable outcomes are {(1,3), (3, 1), (2, 2), (3, 6), (6,3), (4, 5), (5, 4)}
∴ Required probability = \(\frac{7}{36}\)

Probability Class 10 Important Questions Question 10.
Someone is asked to choose a number from 1 to 100. What is the probability of it being a prime number?
Solution:
Total prime numbers between 1 to 100 = 25
∴ P (Prime number) = \(\frac{25}{100}\) = \(\frac{1}{4}\)

Extra Sums Of Probability Class 10 Question 11.
Cards marked with number 3, 4, 5, …., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Solution:
Possible outcomes are 4, 9, 16, 25, 36, 49, i.e., 6.
∴ P (perfect square number) = \(\frac{6}{48}\) or \(\frac{1}{8}\)

Probability Questions Class 10 Question 12.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
Solution:
Number of possible outcomes = 52
Number of red cards and queens = 28
Number of favourable outcomes = 52 – 28 = 24
P (getting neither a red card nor a queen) = \(\frac{24}{52}\) = \(\frac{6}{13}\)

Questions On Probability Class 10 Question 13.
20 tickets, on which numbers 1 to 20 are written, are mixed throughly and then a ticket is drawn at random out of them. Find the probability that the number on the drawn ticket is a multiple of 3 or 7.
Solution:
n(s) = 20, Multiples of 3 or 7, A: {3, 6, 9, 12, 15, 18, 7, 14), n(A) = 8
∴ Required probability = \(\frac{8}{20}\) or \(\frac{2}{5}\)

Probability Important Questions Class 10 Question 14.
A number is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What will be the probability that square of this number is less then or equal to 1?
Solution:
Favourable outcomes are -1, 0, 1 = 3
Total outcomes = 7
∴ Required probability = \(\frac{3}{7}\)

Probability Class 10 Extra Questions Short Answer Type 1

Class 10 Maths Ch 15 Extra Questions Question 1.
Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is a prime number.
Solution:
Product of the number on the dice is prime number, i.e., 2, 3, 5.
The possible ways are, (1, 2), (2, 1), (1, 3), (3, 1), (5, 1), (1, 5)
So, number of possible ways = 6
∴ Required probability = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Class 10th Probability Extra Questions Question 2.
Find the probability that a number selected from the numbers 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.
Solution:
Total prime numbers from 1 to 25 = 9.
∴ Non-prime numbers from 1 to 25 = 25 – 9 = 16.
⇒ P (non-prime number) = \(\frac{16}{25}\)

Probability Class 10 Important Questions 2020 Question 3.
One card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is an ace and black.
Solution:
Number of black aces in a pack of cards = 2
∴ P (an ace and black card) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Ch 15 Maths Class 10 Extra Questions Question 4.
A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Solution:
Let E be the event card drawn is neither an ace nor a king.
Then, the number of outcomes favourable to the event E = 44 (4 kings and 4 aces are not there)
∴ P(E) = \(\frac{44}{52}\) = \(\frac{11}{13}\)

Probability Class 10 Questions Question 5.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?
Solution:
(i) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out an orange flavoured candy is an impossible event. So, its probability is 0.

(ii) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out lemon flavoured candies is certain event. So, its probability is 1.

Question 6.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Here, total number of pens = 132 + 12 = 144
∴ Total number of elementary outcomes = 144
Now, favourable number of elementary events = 132
∴ Probability that a pen taken out is good one = \(\frac{132}{144}\) = \(\frac{11}{12}\)

Question 7.
Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta’s winning the match is 0.62. What is the probability of Reshma’s winning the match?
Solution:
Let S and R denote the events that Sangeeta and Reshma wins the match, respectively.
The probability of Sangeeta’s winning = P(S) = 0.62
As the events R and S are complementary
∴ The probability of Reshma’s winning = P(R) = 1 – P(S)
= 1 – 0.62 = 0.38.

Question 8.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
The total number of elementary events associated with random experiment of throwing a die is 6.
(i) Let E be the event of getting a letter A.
∴ Favourable number of elementary events = 2
∴ P(E) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(ii) Let E be the event of getting a letter D.
∴ Favourable number of elementary events = 1
∴ P(E) = \(\frac{1}{6}\)

Question 9.
A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither a red card nor a black king.
Solution:
Let E be the event card drawn is neither a red card nor a black king’
The number of outcomes favourable to the event E = 24 (26 red cards and 2 black kings are not there, so 52 – 28 = 24)
∴ P(E) = \(\frac{24}{52}\) = \(\frac{16}{13}\)

Question 10.
Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken out at random from the box. Find the probability that the drawn bulb is not defective.
Solution:
Total number of bulbs in the box = 400
Total number of defective bulbs in the box = 15
Total number of non-defective bulbs in the box = 400 – 15 = 385

Question 11.
Rahim tosses two different coins simultaneously. Find the probability of getting at least one
tail.
Solution:
The sample space is {HH, HT, TH, TT}
Total number of outcomes = 4
Outcomes for getting at least one tail is {HT, TH, TT}
Number of favourable outcomes = 3

= \(\frac{3}{4}\)

Probability Class 10 Extra Questions Short Answer Type 2

Question 1.
Harpreet tosses two different coins simultaneously (say, one is of 1 and other of 2). What is the probability that she gets at least one head?
Solution:
When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T) which are all equally likely. Here (H, H) means head up on the first coin (say on ₹ 1) and head up on the second coin (₹ 2). Similarly (H, T) means head up on the first coin and tail up on the second coin and so on.
The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) and (T, H). So, the number of outcomes favourable to E is 3.
Therefore, P(E) = \(\frac{3}{4}\)
i.e., the probability that Harpreet gets at least one head is \(\frac{3}{4}\).

Question 2.
A game consists of tossing a one-rupee coin 3 times and noting the outcome each time. Ramesh wins the game if all the tosses give the same result (i.e. three heads or three tails) and loses otherwise. Find the probability of Ramesh losing the game.
Solution:
The outcomes associated with this experiment are given by
HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
∴ Total number of possible outcomes = 8
Now, Ramesh will lose the game if he gets
HHT, HTH, THH, TTH, THT, HTT
∴ Favourable number of events = 6
∴ Probability that he lose the game = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 3.
Three unbiased coins are tossed together. Find the probability of getting:
(i) all heads.
(ii) exactly two heads.
(iii) exactly one head.
(iv) at least two heads.
(v) at least two tails
Solution:
Elementary events associated to random experiment of tossing three coins are
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
∴ Total number of elementary events = 8

(i) The event “getting all heads” is said to occur, if the elementary event HHH occurs, i.e., HHH is an outcome.
∴ Favourable number of elementary events = 1
Hence, required probability = \(\frac{1}{8}\)

(ii) The event “getting two heads” will occur, if one of the elementary events HHT, THH, HTH occurs.
∴ Favourable number of elementary events = 3
Hence, required probability = \(\frac{3}{8}\)

(iii) The event of “getting one head”, when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH, occurs.
Favourable number of elementary events = 3
Hence, required probability = \(\frac{3}{8}\)

(iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that
the event “getting at least two heads” occurs.
∴ Favourable number of elementary events = 4
honom 4 Hence, required probability = \(\frac{4}{8}\) = \(\frac{1}{2}\)

(v) Similar as (iv) P (getting at least two tails) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 4.
A die is thrown once. Find the probability of getting:
(i) a prime number.
(ii) a number lying between 2 and 6.
(iii) an odd number.
Solution:
We have, the total number of possible outcomes associated with the random experiment of throwing a die is 6 (i.e., 1, 2, 3, 4, 5, 6).
(i) Let E denotes the event of getting a prime number.
So, favourable number of outcomes = 3 (i.e., 2, 3, 5)
∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
(ii) Let E be the event of getting a number lying between 2 and 6.
∴ Favourable number of elementary events (outcomes) = 3 (i.e., 3, 4, 5)
∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
(iii) Let E be the event of getting an odd number.
∴ Favourable number of elementary events = 3 (i.e., 1, 3, 5)
∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 5.
Suppose we throw a die once.
(i) What is the probability of getting a number greater than 4?
(ii) What is the probability of getting a number less than or equal to 4?
Solution:
(i) Here, let E be the event getting a number greater than 4′. The number of possible outcomes are six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is 2. So,
P(E) = P (number greater than 4) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(ii) Let F be the event ‘getting a number less than or equal to 4’.
Number of possible outcomes = 6
Outcomes favourable to the event F are 1, 2, 3, 4.
So, the number of outcomes favourable to F is 4.
Therefore, P(F) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Question 6.
One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will:
(i) be an ace.
(ii) not be an ace.
Solution:
Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck. Let E be the event ‘the card is an ace’.
The number of outcomes favourable to E = 4.
The number of possible outcomes = 52
Therefore, P(E) = \(\frac{4}{52}\) = \(\frac{1}{3}\).
(ii) Let Ē be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event Ē = 52 – 4 = 48.
The number of possible outcomes = 52.
Therefore, P(Ē) = \(\frac{48}{52}\) = \(\frac{12}{13}\)

Question 7.
Five cards – the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Here, the total number of possible outcomes = 5.
(i) Since, there is only one queen
∴ Favourable number of elementary events = 1
∴ Probability of getting the card of queen = \(\frac{1}{5}\)

(ii) Now, the total number of possible outcomes = 4.
(a) Since, there is only one ace
∴ Favourable number of elementary events = 1
∴ Probability of getting an ace card = \(\frac{1}{4}\)

(b) Since, there is no queen (as queen is put aside)
∴ Favourable number of elementary events = 0
∴ Probability of getting a queen = \(\frac{0}{4}\)

Question 8.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white? (iii) not green?
Solution:
Here, total number of marbles = 17
∴ Total number of possible outcomes = 17.
(i) Since, there are 5 red marbles in the box.
∴ Favourable number of elementary events = 5
∴ Probability of getting red marble = \(\frac{5}{17}\)

(ii) Since, there are 8 white marbles in the box.
∴ Favourable number of elementary events = 8
∴ Probability of getting white marble = \(\frac{8}{17}\)

(iii) Since, there are 5 + 8 = 13 marbles which are not green in the box.
∴ Favourable number of elementary events = 13
∴ Probability of not getting a green marble = \(\frac{13}{17}\)

Question 9.
A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) red.
(ii) black or white.
(iii) not black.
Solution:
Total number of balls = 5 + 7 + 3 = 15
Number of red balls = 7, Number of black or white = 5 + 3 = 8 balls
Number of not black = 7 + 3 = 10 balls

(1) P (red ball) = \(\frac{7}{15}\)
(ii) P (black or white ball) = \(\frac{8}{15}\)
(iii) P (not black ball) = \(\frac{10}{15}\) = \(\frac{2}{2}\)

Question 10.
A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is: (i) red or white. (ii) not black. (iii) neither white nor black.
Solution:
Total number of balls = 5 + 8 + 7 = 20
(i) P (red or white) = \(\frac{5+8}{20}\) = \(\frac{13}{20}\)
(ii) P (not black) = 1 – P (black) = 1 – \(\frac{7}{20}\) = \(\frac{13}{20}\)
(iii) P (neither white nor black) = P (Red balls) = \(\frac{5}{20}\) = \(\frac{1}{4}\)

Question 11.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let E be the event of having the same birthday.
Therefore, Ē is the event of not having the same birthday.
i.e., P (Ē) = 0.992 (Given)
Now, we have
P(E) + P(Ē) = 1 ⇒ P(E) = 1 – P(E) = 1 -0.992 = 0.008.

Question 12.
1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Saket has purchased one lottery ticket, what is the probability of winning a prize ?
Solution:
Out of 1000 lottery tickets, one ticket can be chosen in 1000 ways.
∴ Total number of elementary events = 1000
It is given that there are 5 prizes on these 1000 tickets.
Therefore, number of ways of selecting a prize ticket = 5 .
∴ Favourable number of elementary events
Hence, P (Winning a prize) = \(\frac{5}{1000}\) = \(\frac{1}{200}\)

Question 13.
In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice ? (ii) a total of 9 or 11 ?
Solution:
Total outcomes = 36
(i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5) (5, 2) (5, 3) (5, 5) i.e., 9 outcomes.
P (a prime number on each die) = \(\frac{9}{36}\) or \(\frac{1}{4}\)
(ii) Favourable outcomes are (3,6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e., 6 outcomes
P (a total of 9 or 11) = \(\frac{6}{36}\) or \(\frac{1}{6}\)

Question 14.
Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd numbers.
Solution:
Total number of outcomes = 36
(i) Favourable outcomes are
(1, 1,) (1, 2) (1, 3) (1, 4) (1,5) (2, 1) (2, 2) (2, 3)
(2, 4) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1) i.e., 15
∴ P (sum less than 7) = \(\frac{15}{36}\) or \(\frac{5}{12}\)
(ii) Favourable outcomes are
(1, 1) (1, 2) (1,3) (1, 4) (1,5) (1,6) (2, 1) (2, 2) (2,3)
(2, 4) (2,5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3,5) (4, 1)
(4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (6, 1) (6, 2) i.e., 25
∴ P (product less than 16) = \(\frac{25}{36}\)
(iii) Favourable outcomes are
(1, 1) (3, 3) (5, 5) 1.e, 3
∴ (doublet of odd number) = \(\frac{3}{36}\) or \(\frac{1}{12}\)
3 or 1

Probability Class 10 Extra Questions Long Answer Type

Question 1.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red colour.
(ii) a face card.
(iii) a red face card.
(iv) the jack of hearts.
(v) a spade.
(vi) the queen of diamonds.
Solution:
Here, total number of possible outcomes = 52
(i) As we know that there are two suits of red card, i.e., diamond and heart and each suit contains one king.
∴ Favourable number of outcomes = 2
∴ Probability of getting a king of red colour = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(ii) As we know that kings, queens and jacks are called face cards. Therefore, there are 12 face cards.
∴ Favourable number of elementary events = 12
∴ Probability of getting a face card = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(iii) As we know there are two suits of red cards, i.e., diamond and heart and each suit contains 3 face cards.
∴ Favourable number of elementary events = 2 × 3 = 6
∴ Probability of getting red face card = \(\frac{6}{52}\) = \(\frac{3}{26}\)

(iv) Since, there is only one jack of hearts.
∴ Favourable number of elementary events = 1
∴ Probability of getting the jack of heart = \(\frac{1}{52}\)

(v) Since, there are 13 cards of spade.
∴ Favourable number of elementary events = 13
∴ Probability of getting a spade = \(\frac{13}{52}\) = \(\frac{1}{4}\)

(vi) Since, there is only one queen of diamonds.
∴ Favourable number of outcomes (elementary events) = 1
∴ Probability of getting a queen of diamond = \(\frac{1}{52}\)

Question 2.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is:
(i) an ace.
(ii) red.
(iii) either red or king.
(iv) red and a king.
(v) a face card.
(vi) a red face card.
(vii) “2′ of spades.
(viii) ’10’ of a black suit.
Solution:
Out of 52 cards, one card can be drawn in 52 ways.
So, total number of elementary events = 52
(i) There are four ace cards in a pack of 52 cards. So, one ace can be chosen in 4 ways.
∴ Favourable number of elementary events = 4
Hence, required probability = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(ii) There are 26 red cards in a pack of 52 cards. Out of 26 red cards, one card can be chosen in 26 ways.
∴ Favourable number of elementary events = 26
Hence, required probability = \(\frac{26}{52}\) = \(\frac{1}{2}\)

(iii) There are 26 red cards, including two red kings, in a pack of 52 playing cards. Also, there are 4 kings, two red and two black. Therefore, card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings).
∴ Favourable number of elementary events = 28
Hence, required probability = \(\frac{28}{52}\) = \(\frac{7}{13}\)

(iv) A card drawn will be red as well as king, if it is a red king. There are 2 red kings in a pack of 52 playing cards.
∴ Favourable number of elementary events = 2
Hence, required probability = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(v) In a deck of 52 cards: kings, queens, and jacks are called face cards. Thus, there are 12 face cards. So, one face card can be chosen in 12 ways.
Favourable number of elementary events = 12
Hence, required probability = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(vi) There are 6 red face cards 3 each from diamonds and hearts. Out of these 6 red face cards, one card can be chosen in 6 ways.
∴ Favourable number of elementary events = 6
Hence, required probability = \(\frac{6}{52}\) = \(\frac{3}{26}\)

(vii) There is only one ‘2’ of spades.
∴ Favourable number of elementary events = 1 Hence, required probability = 2

(viii) There are two suits of black cards viz. spades and clubs. Each suit contains one card bearing number 10.
∴ Favourable number of elementary events = 2
Hence, required probability = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 3.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8 see Fig, and these are equally likely outcomes. What is the probability that it will point at: (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Solution:

Here, total number of elementary events (possible outcomes) = 8
(i) We have only one ‘P’ on the spining plant.
∴ Favourable number of outcomes = 1
Hence, the probability that arrow points at 8 = \(\frac{1}{26}\).

(ii) We have four odd points (i.e., 1, 3, 5 and 7)
∴ Favourable number of outcomes = 4
∴ Probability that arrow points at an odd number = \(\frac{4}{8}\) = \(\frac{1}{2}\)

(iii) We have 6 numbers greater than 2, i.e., 3, 4, 5, 6, 7 and 8.
Therefore, favourable number of outcomes = 6
∴ Probability that arrow points at a number greater than 2 = \(\frac{6}{8}\) = \(\frac{3}{4}\)

(iv) We have 8 numbers less than 9, i.e, 1, 2, 3, … 8.
∴ Favourable number of outcomes = 8
∴ Probability that arrow points at a number less than 9 = \(\frac{8}{8}\) = 1.

Question 4.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is: (i) 8? (ii) 13? (iii) less than or equal to 12?
Solution:
When the blue die shows ‘l’, the grey die could show any one of the numbers 1, 2, 3, 4, 5, 6.
The same is true when the blue die shows ‘2’, ‘3’, ‘4’, ‘5’ or ‘6’. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die. So, the number of possible outcomes = 6 × 6 = 36.

(i) The outcomes favourable to the event the sum of the two numbers is 8′ denoted by E, are :
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see figure)
i.e., the number of outcomes favourable to E = 5.
Hence, P(E) = \(\frac{5}{36}\)

(ii) As you can see from figure, there is no outcome favourable to the event F, ‘the sum of two numbers is 13’.
So, P(F) = \(\frac{0}{36}\) = 0

(iii) As you can see from figure, all the outcomes are favourable to the event G, ‘sum of two numbers ≤ 12.
So, P(G) = \(\frac{36}{36}\) = 1.

Question 5.
A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards throughly. Find the probability that the number on the drawn card is:
(i) an odd number.
(ii) a multiple of 5.
(iii) a perfect square.
(iv) an even prime number.
Solution:
Total number of cards = 49
Total number of outcomes = 49
(i) Odd number
Favourable outcomes : 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49
Number of favourable outcomes = 25

Question 6.
All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a:
(i) face card.
(ii) red card.
(iii) black card.
(iv) king.
Solution:
Cards remaining after removing black face cards = red cards + black cards excluding face cards
= 26 + 20 = 46
Total number of possible outcomes = 46
(i) Face Card
Favourable outcomes: 6 red face cards (king, queen and jack of diamond and heart suits)

Question 7.
Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is:
(i) an odd number.
(ii) a perfect square number.
(iii) divisible by 5.
(iv) a prime number less than 20.
Solution:
No. of possible outcomes = 60 – 11 + 1 = 50.
(i) An odd number
Favourable outcomes : 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59
No. of favourable outcomes = 25
Number of favourable outcomes

= \(\frac{25}{50}\) = \(\frac{1}{2}\)

(ii) A perfect square number
Favourable outcomes : 16, 25, 36, 49
No. of favourable outcomes = 04

= \(\frac{4}{50}\) = \(\frac{2}{25}\)

(iii) Divisible by 5
Favourable outcomes : 15, 20, 25, 30, 35, 40, 45, 50, 55, 60
No. of favourable outcomes = 10

= \(\frac{10}{50}\) = \(\frac{1}{5}\)

(iv) A prime number less than 20
Favourable outcomes : 11, 13, 17, 19
No. of favourable outcomes = 4

= \(\frac{4}{50}\) = \(\frac{2}{25}\)

Question 8.
A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.
Solution:
x can be any one of 1, 2, 3 or 4.
y can be any one of 1, 4, 9 or 16
Total number of cases of product of x and y = 16
Product less than 16 = (1 × 1, 1 × 4, 1 × 9, 2 × 1, 2 × 4, 3 × 1, 3 × 4, 4 × 1)
Number of cases, where product is less than 16 = 8
∴ Required probability = \(\frac{8}{16}\) or \(\frac{1}{2}\)

Question 9.
In Fig, shown a disc on which a player spins an arrow twice. The function \(\frac{a}{b}\) is formed, where ‘a’ is the number of sector on which arrow stops on the first spin and ‘b’ is the number of the sector in which the arrow stops on second spin. On each spin, each sector has equal chance of selection by the arrow. Find the probability that the fraction \(\frac{a}{b}\) > 1.
Solution:

For alb > 1, when a = 1, b can not take any value,
a = 2, b can take 1 value,
a = 3, b can take 2 values,
a = 4, b can take 3 values,
a = 5, b can take 4 values,
a = 6, b can take 5 values.
Total possible outcomes = 36

Question 10.
Two different dice are thrown together. Find the probability that the numbers obtained have
(i) even sum, and (ii) even product.
Solution:
Total number of outcomes = 36[(1, 1), (1, 2) … (6,6)]
Number of outcomes when sum is even = 18 [(1, 1), (1, 3) …(6, 6)]
Number of outcomes when product is even = 27 [(1, 2), (1, 4) … (6,6)]

Probability Class 10 Extra Questions HOTS

Question 1.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
Solution:
(i) Favourable outcomes are
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3,6), (4, 1), (4,2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)]
Total outcomes = 36
∴ Required probability = \(\frac{25}{36}\)
(ii) Probability that 5 will come atleast once = 1 – P (5 will not come up either time)
= 1 – \(\frac{25}{36}\) = \(\frac{11}{36}\)

Question 2.
Find the probability that in a leap year there will be 53 Tuesdays.
Solution:
Leap year = 366 days = (52 × 7 + 2) days = 52 weeks and 2 days.
Thus, a leap year always has 52 Tuesdays. The remaining 2 days can be:

1. Sunday and Monday
2. Monday and Tuesday
3. Tuesday and Wednesday
4. Wednesday and Thursday
5. Thursday and Friday
6. Friday and Saturday
7. Saturday and Sunday

Out of these 7 cases, we have Tuesdays in two cases
∴ P(53 Tuesdays) = \(\frac{2}{7}\)

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.
Solution:
Let there be x blue balls in the bag.
∴ Total number of balls in the bag = (5 + x)

Hence, there are 15 blue balls in the bag.

Question 4.
Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why?
Solution:
Apoorv throws two dice once.
So, total number of outcomes, n(S) = 36.
Number of outcomes for getting product 36,
n(E₂) = 1 [(6 × 6)]
∴ Probability for Apoorv getting the number 36

Also, Peehu throw one die.
So, total number of outcomes n(S) = 6
Number of outcomes for getting square of a number as 36.
n(E₂) = 1 (∵ 6² = 36)
∴ Probability for Peehu getting the number 36

Hence, Peehu has better chance of getting the number 36.

Check the below NCERT MCQ Questions for Class 10 History Chapter 6 Extra Questions and Answers Work-Life and Leisure with Answers Pdf free download. https://ncertmcq.com/extra-questions-for-class-10-social-science/

Work-Life and Leisure Class 10 Extra Questions History Chapter 6

Class 10 History Chapter 6 Questions And Answers Question 1.
From where do the migrants usually come in the cities. Mention one example.
Answer:
Usually, the migrants come from nearby rural areas. In 1851, more than three fourth of the adults in Manchester (Britain) were people from the nearby rural region.

Writing And City Life Extra Question And Answer Question 2.
Name the five types of industries which grew in London in the 19th century.
Answer:

1. Clothing and footwear,
2. wood and furniture,
3. metal and engineering,
4. printing and stationary precision products such as surgical instruments.

Class 10 History Chapter 6 Questions And Answers In Bengali Question 3.
Who is a philanthropist?
Answer:
Someone who works for social upliftment and charity donating time and money is known as philanthropist.

Class 10 History Chapter 6 Mcq With Answers Question 4.
What is a metropolis?
Answer:
Metropolis is a large densely city of a country.

History Class 10 Chapter 6 Question Answer Question 5.
Note the second largest city of the world.
Answer:
Mumbai, also an important city of the Indian subcondiment.

Ncert Solutions For Class 10 Social Science History Chapter 6 Question 6.
How can you define urbanisation?
Answer:
Urbanisation is the development of a city or town terms” of trade and expanding of population.

Work, Life And Leisure Class 10 Solutions Question 7.
When was the underground railway opened in London and between which two places?
Answer:
The underground railway was opened on January 10, 1863, and between Paddington and Farrington street in London.

Class 10 History Chapter 6 Question 8.
What is asphyxiation?
Answer:
Asphyxiation is suffocation due to lack of oxygen supply.

Question 9.
Give two reasons which enabled large number of people to live outside London and travel to work.
Answer:

1. Better planned suburbs and
2. a good railway network.

Question 10.
What do you mean by individualism?
Answer:
Individualism is a theory which promotes liberty rights, or independent action of the individual rather than of the”community.

Question 11.
What and when were the entertainment resorts in the 19th century London?
Answer:
Pleasure gardens became entertainment resorts in the 19th century London providing facilities such as sports and entertainments and refreshments for the well-to-do.

Question 12.
What was the Bloody Sunday of November 1887?
Answer:
A riot brutely suppressed by the Police in 1887 is known as Bloody Sunday in London.

Question 13.
Why is a large city an opportunity?
Answer:
A large city is an opportunity because it provides employment, as it did with the industrialization coming in.

Question 14.
Name the three Presidencies and their cities in the colonial India.
Answer:
The three Presidencies were: Bengal, Bombay and Madras, and three Presidencies were Calcutta, Bombay, and Madras.

Question 15.
When and where was the first cotton textile mill established in India?
Answer:
The first cotton textile till was established in Bombay in 1854.

Question 16.
When and why was the women mill workforce drop in Bombay?
Answer:
Between 1919 and 1926, the women mill workforce was 27% of the total workers. It dropped to 10% by 1930, due to machines or men taking on the jobs.

Question 17.
What is ‘mill village’?
Answer:
Place of living by the workers near their place of work.

Question 18.
What do you mean by chawls in reference to Bombay?
Answer:
Chawls, in reference to Bombay, were multi-storeyed structures built in the native parts of the city of Bombay.

Question 19.
What were the three major groups of people in Singapore in the early days of its formation?
Answer:
Do yourself.

Question 20.
Give an account of expansion of the city of London.
Answer:
By 1750, one out of every nine people of England and Wales lived in London. It was a colossal city with a population of about 675,000. Over the nineteenth century, London continued to expand. Its population multiplied fourfold in the 70 years between 1810 and 1880, increasing from 1 million to about 4 million.

Question 21.
Why did the poor die of the lack of housing accommodation in late 19lh century London?
Answer:
The lack of proper housing becomes a cause for early death and of poverty. Charles Booth, a Liverpool shipowner writes that as many as a million Londoners (about one-fifth of the population of London at the time were very poor and were expected to live only up to an average age of 29 (compared to the average lile expectancy of 55 among the gentry and the middle class).

These people were more than likely to die in a ‘workhouse, hospital or lunatic asylum. London, he concluded ‘needed the rebuilding of at least 400,000 rooms to house its poorest citizens.

Question 22.
Why were people afraid of travelling underground when the underground railway began functioning?
Answer:
The people, at first, were afraid of travelling the unground railways. This is what one newspaper reader reported: “The compartment in which I sat was filled with passengers who were smoking pipes. The atmosphere was a mixture of sulphur, coal dust and foul fumes from the gas lamps above so that by the time we reached Moorgate, I was near dead of asphyxiation and heat should think these underground railways must soon be discontinued for they eat health.

Question 23.
How do you explain that the curly industrial life was largely male-dominated?
Answer:
With industrialisation and urbanisation, men and women did not have equal access to this new urban space. As women lost their industrial jobs and conservative people tailed against their presence in public spaces, women were forced to withdraw into their homes.

The public space became increasingly a male preserve and the domestic sphere was seen as the proper place for women. Most political movements of the nineteenth century, such as Chartism (a movement demanding the role for all adult males). and the 10-hour movement (limiting hours of work in factories), mobilised large numbers of men.

Question 24.
Give examples of leisure and entertainment activities as cities grew.
Answer:
As cities grew, there also grew leisure, art, and entertainment. In London, libraries and museums came up: seaside holidaying was resorted; music halls beginning functioning. When the entry to the British Museum Library became free in 1810, visitors swamped the museum their number jumped to 127,643 in 1824-25, shooting up to 25,901 by 1846. Music-halls were popular among the lower classes, and, by the early twentieth century, cinema became the greal mass entertainment for mixed audiences.

British industrial workers were increasingly encouraged to spend their holidays by the sea, so as to derive the benefits of the sun and bracing winds Over 1 million British people went to the seaside at Blackpool in 1883, by 1939 their numbers had gone up to 7 million,

Question 25.
Give a brief history of the city of Bombay.
Answer:
In the 17th century, Bombay was a group of seven islands under Portuguese control. In 1661, control of the islands passed into British hands after the marriage of Britain’s King Charles. It to the Portuguese princess. The East India Company quickly shifted its base from Surat, its principal western port to Bombay.

At first, Bombay was the major outlet for cotton textiles from Gujarat. Later, in the nineteenth century, the city functioned as a port through which large quantities of raw materials such as cotton and opium would pass. Gradually, it also became an important administrative centre in western India, and then, by the end of the nineteenth century, a major industrial centre.

Question 26.
When was the Bombay improvement Trust established? What was the object of the Rent Act of 1918?
Answer:
The city of Bombay Improvement Trust was established in 1898. It focused on clearing poorer houses out of the city centre, By 1918, Trust schemes had deprived 64,000 people of their homes, but only 14,000 were rehoused. In 1918, a Rent Act was passed to keep rents reasonable, but it had the opposite effect of producing a severe housing crisis since landlords withdrew houses from the market.

Question 27.
What was the contribution of the Bombay Port Trust in reclaim nation project?
Answer:
A successful reclamation project was undertaken by the Bombay Port Trust, which built a dry dock between 1914 and 1918 and used the excavated earth to create the 22-acre Ballard Estate. Subsequently, the famous Marine Drive or Bombay was developed.

Question 28.
Give a brief account of the life of the working women in numerous jobs.
Answer:
Factories employed large numbers of women in the late eighteenth and early nineteenth centuries. With Technological developments, women gradually lost their industrial jobs and were forced to work within households.

The 1861 census recorded a quarter of a million domestic servants in London, of whom the vast majority were women, many of them recent migrants A large number of women used their homes to increase family income by taking in lodgers or through such activities as tailoring, washing or matchbox making.

However, there was a change once again in the twentieth century. As women got employment in wartime industries and offices, they withdrew from cosmetic service.

Question 29.
Mention a variety of steps which were taken to clean up London during 19th-20th centuries.
Answer:
A variety of steps were taken to clean up London.

• Attempts were made to decongest localities, green the open spaces, reduce pollution and landscape the city.
• Large blocks of apartments were built, akin to those in Berlin and new York-cities which had similar housing problems.
• Rent control was introduced in Britain during the First World War to case the impact of a severe housing shortage.
• (iv) Many wealthy residents of London were able to afford a holiday home in the countryside.
• Architect and planner Ebenezer Howard developed the principle of the Garden City, a pleasant space full of plants and trees, where people would both live and work. Following Howard’s ideas, Raymond Unwin and Barry Parker designed the garden city.
• There were common grade 1 spaces, beautiful views and great attention to detail.
• Between the two world wars (1919-39) the responsibility for housing the working classes was accepted by the British state and a million houses, most of them single-family cottages, were built by local authorities.

Question 30.
What are Chawls of Bombay? Give an account of life and living in a Chawl.
Answer:
Chawls are multi-storeyed structures which had been built in 1860s in the town occupied by the indigenous natives, mostly owned by landlords and merchants. Each chawl was divided into smaller one-room tenements which had no private toilets.

Many families could reside at a time in a lenement The Census of 1901 reported that the mass of the island’s population or 80 percent of the total resides in tenements of one room, the average number of occupants lies between 4 and 5. High rents forced workers to share homes, eiiter with relatives or caste fellows who were streaming into the city.

People had to keep the windows of their rooms closed even in humid weather due to the close proximity of filthy gutters, privies, buffalo stables etc.’Yet, though water was scarce, and people often quarrelled every morning for a turn at the tap observers found that houses were kept quite clean.

The homes being shall, streets and neighbourhoods were used for a variety of activities such as cooking, washing and sleeping Liquor shops and akharas came up in any emply spot. Streets were also used for different types of leisure activities.

Chawls were also the place for exchange of news about jobs, strikes, riots or demonstrations.
Caste and family groups in the mill neighbourhoods were headed by someone who was similar to a village headman. Sometimes, the jobber in the mills could be the local neighbourhood leader He settled disputes, organised food supplies, or arranged informal credit.

Question 31.
Give a brief description of how Singapore developed as a planned nation.
Answer:
Singapore is a successful, rich, and well-planned city, a model for city planning worldwide. Yet the city’s rise to this status is quite recent. Until 1965, Singapore, though an important port, shared all the problems of other Asian cities, Planning was known in Singapore since 1822, but benefited only the small community of white people who ruled Singapore.

For the majority of its inhabitants, there was overcrowding, lack of sanitation, poor housing and poverty. All this changed after the city became an independent nation in 1965 under the leadership of Lee Kuan Yew, President of the People’s Action Party.

A massive housing and development programme was undertaken and it completely altered the face of the island nation. Through a programme of total planning which left nothing to chance, every inch of the island’s territory was controlled in its use. The government itself wen popular support by providing nearly 85 percent of the population with ownership housing of good quality.

The tall housing blocks, which were well ventilated and serviced, were examples of good physical planning. But the buildings also redesigned social life: crime was reduced through external corridors, the aged were housed alongside their families, ‘void decks’ or empty floors were provided in all buildings for community activities.

Objective Type Questions

1. Choose the most appropriate alternative:

Question 1.
Metapolisis
(a) large but not a densely city
(b) a densely but a small city
(c) large as well as densely city
(d) none of the above
Answer:
(c) large as well as densely city

Question 2.
The capital of Britain is:
(a) Manchester
(b) London
(c) Liverpool
(d) Blackpool
Answer:
(b) London

Question 3.
Baron Haussmann was a French
(a) poet
(b) legislator
(c) architect
(d) shoemaker
Answer:
(c) architect

Question 4.
Dadasahab Phalke made the following movie :
(a) Raja Harishchandra
(b) Raja Ram Mohan Roy
(c) Raja jai Singh
(d) Raja and Rani
Answer:
(a) Raja Harishchandra

2. Choose true (✓) or False (✗):

Question 1.
Bombay is India’s film capital
Answer:
(✓)

Question 2.
Ismat Chughtai and Saadat Hasan Manto were associated with tea trade in India
Answer:
(✗)

Question 3.
Haussmann was an architect.
Answer:
(✓)

Question 4.
Individualism is a theory of community.
Answer:
(✗).

Extra Questions for Class 10 Social Science

Check the below NCERT MCQ Questions for Class 10 History Chapter 7 Extra Questions and Answers Print Culture and the Modern World Pdf free download. https://ncertmcq.com/extra-questions-for-class-10-social-science/

Print Culture and the Modern World Class 10 Extra Questions History Chapter 7

Class 10 History Chapter 7 Questions And Answers Question 1.
What is calligraphy?
Answer:
Calligraphy is the art of beautiful and stylised writings.

Print Culture And The Modern World Extra Questions Question 2.
Where did the earliest print technology develop in the world?
Answer:
The earliest print technology developed in China, Japan and Korea; it was, their, hand printing, around 594 AD.

Print Culture And The Modern World Class 10 Important Questions And Answers Question 3.
Name the explorer who brought with him the print technology to Italy.
Answer:
Marco Polo, in 1295.

Print Culture And The Modern World Question Answers Pdf Question 4.
Who had developed the first known printing press and when?
Answer:
Johann Gutenberg, in 1430.

Class 10 History Print Culture And Modern World Notes Question 5.
Who was Martin Luther?
Answer:
Martin Luther was a Protestant reformer. He wrote Ninety Five Thesis in 1517, criticizing the practice and rituals of the Roman Catholic church.

Print Culture And The Modern World Class 10 Questions And Answers Question 6.
Why was Luther deeply grateful to printing?
Answer:
Printing, for Martin Luther, was one of the greatest gift of God to the people.

Class 10 History Chapter 7 Mcq With Answers Question 7.
Why did the Roman church begin maintaining an Index of Prohibited Books from 1558 onward?
Answer:
The Roman church was troubled by the effects of the popular reading questioning it. So it maintained such an index from 1558 onward.

Class 12 History Chapter 7 Extra Questions Question 8.
What were Priliotheque Bleue ?
Answer:
Peilictheque Bleue were low-priced small books in France.

Class 12 History Chapter 7 Extra Questions Question 9.
Whose writings, in France, were widely printed and read during the eighteenth century?
Answer:
Voltaire, and Rousseau.

Print Culture And The Modern World Important Questions Question 10.
Name a socialist whose discoveries, when published, influenced scientifically-minded readers.
Answer:
Isaac Newton.

Print Culture Class 10 Important Questions Question 11.
Whose novels, during the mid 18th century, had provided the basis for opposing despotion?
Answer:
Louise Sebastien Mercier’s novels destroyed the basis of despotism. He once proclaimed: “Tremble, therefore, tyrants of the world! Tremble before the virtual writer.

Print Culture And The Modern World Pdf Question 12.
Which magazines were especially meant for women?
Answer:
Penny magazines.

Ch 7 Science Class 10 Extra Questions Question 13.
How were the earlier manuscripts written in India?
Answer:
The earlier manuscripts were handwritten and in languages such as Sanskrit Persian, Arabic, and also in Vernacular language.

Class 7 Science Chapter 7 Extra Questions Question 14.
Who brought the printing technology to India?
Answer:
The Portuguese.

Class 6 Science Ch 7 Extra Questions Question 15.
Who brought out the weekly Bengal Gazzette?
Answer:
Gangadhar Bhattacharya, a close associate of Raja Ram Mohan Roy.

Question 16.
When was Tulsidas’s Ramcharitmas brought out in the printed form?
Answer:
In 1810, from Calcutta.

Question 17.
Name the two presses of mid-19lh century India which published religious texts.
Answer:

1. Naval Kishore Press (Lucknow).
2. Seri Venkateshwar Press (Bombay).

Question 18.
Give two advantages of the print culture.
Answer:

1. It stimulates the publication of conflicting opinions among communities;
2. It connected communities and people in different parts of India.

Question 19.
Name the book Altaf Husain Ali I wrote in 1874?
Answer:
Majalis Un Nissu (Assemblies of Women).

Question 20.
Which book did Maulana Ashraf Ali write in 1905?
Answer:
Bihishit Zewar (The ornament of Paradise).

Question 21.
Name the first full-length autobiography published in the Bengali language.
Answer:
Amar Jiban by Rashundari Debi.

Question 22.
Who wrote Istri Dharam Vichae?
Answer:
Ram Chaddha.

Question 23.
What does Battrola signify?
Answer:
Battrala signifies the area developed to the printing of popular books. It is located in central Calcutta.

Question 24.
For what art form is Kitagawa Utamaro known? 1
Answer:
Kitagawa Utamaro, born in Edo in 1753, was widely known for his contributions to an art form called ukiyo (‘pictures of the floating world’) or depiction of ordinary human experiences, especially urban ones. These prints travelled to contemporary US and Europe and influenced artists like Manet; and Van Gogh.

Question 25.
How did print technology travel from China to Italy and later in other parts of Europe?
Answer:
In 1295, Marco Polo, a great explorer, returned to Italy after many years of exploration in China. Marco Polo brought this knowledge back with him. Now Italians began producing books with woodblocks, and soon the technology spread to other parts of Europe, where luxury editions were still handwritten on very expensive vellum, meant for aristocratic circles and rich monastic libraries.

Question 26.
What do you know by the print resolution?
Answer:
Print resolution was not just a development, a new way of producing books; it transformed the lives of people, changing their relationship to information and knowledge, and with institutions and authorities. It influenced popular perceptions and opened up new ways of looking at things.

Question 27.
What is reading mania? Explain.
Answer:
Through the seventeenth and eighteenth centuries, literacy rates went up in most parts of Europe. Churches of different denominations set up schools in villages, carrying literacy to peasants and artisans.

By the end of the eighteenth century, in some parts of Europe literacy rates were as high as 60 to 80 per cent. As literacy and schools spread in European countries, there was a virtual reading mania. People wanted books to read and printers produced books in ever-increasing number.

Question 28.
How was the printing press a danger to despotism? What does Mercier say about it?
Answer:
Louis Sebastien Mercier, a novelist in eighteenth-century France, declared: “The printing press is the most powerful engine of progress and public opinion is the force that will sweep despotism away.’ In many of Mercier’s novels, the heroes are transformed by acts of reading.

They devour books, are lost in the world books create, and become enlightened in the process. Convinced of the power of print in bringing enlightenment and destroying the basis of despotism, Mercier proclaimed: Tremble, therefore, tyrants of the world! Treble before the virtual writer.

Question 29.
What do you know about the Grimm Brothers of Germany?
Answer:
The Grimm Brothers in Germany spent years compiling traditional folk tales gathered from peasants. What they collected was edited before the stories were published in a collection in 1812. Anything that was considered unsuitable for children or would appear vulgar to the elites, was not included in the published version.

Question 30.
‘Women were important readers as well as writers’ Explain.
Answer:
Women became important readers as well as writers Penny magazines were especially meant for women, as were manuals teaching proper behaviour and housekeeping. When novels began to be written in the nineteenth-century women were seen as important readers.

Some of the best-known novelists were women: Jane Austen, the Bronte sisters, George Eliot. Their writings became important in defining a new type of woman: a person with will, strength of personality, determination and the power to think.

Question 31.
How did print come to India? Give a brief account.
Answer:
The printing press first came to Goa with Portuguese missionaries in the mid-sixteenth century. Jesuit priests learnt Konkani and printed several tracts. By 1674, about 50 books had been printed in the Konkani and in Kanara languages: Catholic priests printed the first Tamil book in 1579 at Cochin, and in 1713 the first Malayalam book was printed by them. By 1710, Dutch Protestant missionaries had printed 32 Tamil texts, many of them translations of older works.

Question 32.
Which printed technology appeared in 1820s in India?
Answer:
Raja Ram Mohan Roy published Sambd Kaumudi from 1821 and the Hindu orthodoxy commissioned the Samachar Chandrika to oppose his opinions. From 1822. Iwo Persian newspapers were published, Jam-i-Jahu Nama and Samsul Akhbar. In the same year, a Gujarati newspaper, the Bombay Samachar, made its appearance.

Question 33.
What new visual culture was taking place in late 19th century India?
Answer:
By the end of the nineteenth century, a new visual culture was taking shape. With the setting up of an increasing number of printing presses, visual images could be easily reproduced in multiple copies. Painters like Raja Ravi Verma produced images for mass circulation.

Question 34.
How did the print technology development in China?
Answer:
The earliest kind of print technology was developed in China. This was a system of hand printing. From AD 594 onwards, books in China were printed by rubbing paper – also invented there against the inked surface of ‘Woodblocks. As both sides of the thin, porous sheet could not be printed, the traditional Chinese ‘accordion book’ was folded and stitched at the side. Superbly skilled craftsmen could duplicate, with remarkable accuracy the beauty of calligraphy.

The imperial state in China was, for a very long time, the major producer of printed material. China possessed a huge bureaucratic system which recruited its personnel through civil service examinations Textbooks for this examination were printed in vast numbers under the sponsorship of the imperial state.

From the sixteenth century, the number of examination candidates Went up and that increased the volume of print. By the seventeenth century, as urban culture bloomed in China, the uses of print diversified Print was no longer used just by scholar-officials.

Merchants used print in their everyday life, as they collected trade information. Reading increasingly became a leisure activity. From hand-printing, there began mechanical printing. From China, the print technology moved to Europe through the Silk Route.

Question 35.
Did everyone welcome printed books? Why were some people apprehensive?
Answer:
Printed books were both welcomed as well as unwelcomed. Those who disagreed with established authorities could now print and circulate their ideas. Through the printed message, they could persuade people to think differently and move them to action. This had significance in different spheres of life. Not everyone welcomed the printed book and those who did also had fears about it. Many were apprehensive of the effects that the easier access to the printed word and the wider circulation of books, could have on people’s minds.

It was feared that if there was no control over what was printed and read then rebellious and irreligious thoughts might spread. If that happened the authority of valuable literature would be destroyed. Expressed by religious authorities and monarchs, as well as many writers and artists, this anxiety was the basis of widespread criticism of the new printed literature that had began to circulate.

Question 36.
Why do some historians think that the print culture created the basis for the French Revolution? What are the three types of argument? Explain.
Answer:
Many historians have argued that print culture created the conditions within which French Revolution occurred.
Three types of arguments have been usually put forward.

First: print popularized the ideas of the Enlightenment thinkers. Collectively, their writings provided a critical commentary on tradition, superstition and despotism. They argued for the rule of reason rather than custom and demanded that everything be judged through the application of reason and rationality.

They attacked the sacred authority of the Church and the despotic power of the state, thus eroding the legitimacy of a social order based on tradition. The writings of Voltaire and Rousseau were read widely; and those who read these books saw the world through new eyes, eyes that were questioning critical and rational.

Second: Print created a new culture of dialogue and debate. All values, norms and institutions were re-evaluated and discussed by a public that had become aware of the power of reason and recognised the need to question existing ideas and beliefs Within this public culture, new ideas of social revolution came into being.

Third: by the 1780s there was an outpouring of literature that mocked the royalty and criticised their morality. In the process, it raised questions about the existing social order. Cartoons and caricatures typically suggested that the monarchy remained absorbed only in sensual pleasures while the common people suffered immense hardships.

This literature was circulated underground and led to the growth of hostile sentiments against the monarchy. Indeed, print did not directly shape their minds but it did open the possibility of thinking differently.

Question 37.
Bring out newer innovations in the print technology since the eighteenth century.
Answer:
By the late eighteenth century, the press came to be made out of metal. Through the nineteenth century, there were a series of further innovations in printing technology. By the mid-nineteenth century, Richard M. Hoe of New York had perfected the power-driven cylindrical press. This was capable of printing 8,000 sheets per hour. This press was particularly useful for printing newspapers. In the late nineteenth century, the offset press was developed which could print up to six colours at a time.

From the turn of the twentieth century, electrically operated presses accelerated printing operations. A series of other developments followed. Methods of feeding paper improved, the quality of plates became better, automatic paper reels and photoelectric controls of the colour register were introduced. The accumulation of several individual mechanical improvements transformed the appearance of printed texts.

Objective Type Questions

1. Choose the most appropriate alternative:

Question 1.
On what model was the Vernacular Press Act (1878) was passed in India?
(a) American Press Laws
(b) French Press Law’s
(c) Irish Press Laws
(d) Australian Press Laws
Answer:
(c) Irish Press Laws

Question 2.
Istri Dharsun Vichar was written by:
(a) Ram Chhddha
(b) Laxman Chaddha
(c) Bharat Chaddha
(d) None of the above
Answer:
(a) Ram Chhddha

Question 3.
Raja Ravi Varma was:
(a) an architect
(b) a doctor
(c) a painter
(d) a novelist.
Answer:
(c) a painter

Question 4.
Chapbooks were:
(a) low-priced books
(b) high-priced books
(c) cheap books
(d) none of the above
Answer:
(a) low-priced books

2. Choose true (✓) or false (✗) from the following:

Question 1.
Gutenberg was a French artist.
Answer:
(✗)

Question 2.
Print technology owes its origin to Japan.
Answer:
(✗)

Question 3.
Warren Hastings was the Governor-General of Australia.
Answer:
(✗)

Question 4.
Amar Jiban was the first full-length autobiography written in the Assamese language.
Answer:
(✗).

Extra Questions for Class 10 Social Science