Burn Hall Hr. Sec. Schoolburnhallschool.ac.in/wp-content/uploads/2020/05/Class-8...THE BANGLE...

Term 1st Tutorial for Classes 8th ABC - Session 2019-20

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Burn Hall Hr. Sec. School Gupkar Road, Sonwar, Srinagar, Jammu & Kashmir - 190 001

www.burnhallschool.ac.in

SUBJECT – ENGLISH

THE BANGLE-SELLERS BY- SAROJINI NAIDU

Ans 1) The bangle seller is the speaker in the poem. He describes himself and his bangles throughout the poem.

Ans 2) In the first stanza the bangles are described as lustrous, shining loads, rainbow coloured, delicate and bright for happy daughters and happy wives .

Ans3) The similes used in the poem are: like the mountain mist, like newborn leaves, like sunlit corn, like the flame of her marriage fire and like bridal laughter and bridal tear.

Ans 4) Silver and blue, rainbow coloured, sunlit corn, purple gold flecked grey are some of the colours used for the bangles suitable for women of different age groups.

Ans 5) The word ‘Some’ represents different type of bangles that an Indian woman wishes to wear at different stages of her life. Ans 6) These lines have been taken from the second stanza of the poem , ‘The Bangle Sellers’, written by Sarojini Naidu. In these lines the poet says that some bangles are made for the wrists of unmarried women with silver and blue colour just like the mist of the mountains. Word pyramid Answers: 1) Mob 2)Moth 3)Month 4)Mother 5)Morning 6)Mountain

Language work

Fill in the blanks with the words given 1. Bangles are said to be as-rainbow tinted circles of light. 2. Some are like fields of sunlit corn 3. Some bangles are purple and gold flecked grey. 4. People buy these bangles from a bangle-seller.— Prayer for strength By Rabindranath Tagore Q1) Why does the poet want God to strike at his heart? Ans) The poet prays to God to remove the hardheartedness and to fill up his heart with love and passion. He wants God to strike at his heart so that he could bear the joys and sufferings of life.

Ans2) The poet wants strength to bear the joys and sorrows, to make his life fruitful in service, never to disown the poor and never to bend before rude people.

Ans3) You can make your life meaningful by changing the life of your beloved in a fruitful way with sweet results.

Ans4) We should never disown the poor. We should have love and compassion for them. Ans5) ‘Bend my knees’ signifies to resist or oppose disrespectful power and not to bend before them.

Ans6) The poet wants to raise his mind above daily trifles because he wants to overcome things of little value and think about more important things that may be beneficial for the people. Ans7) The poet wants strength to surrender his will before God’s will because his heart will get purified and shall be loved by God.


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LANGUAGE WORK: Prefixes:

1. Disagree 2. Dishonour 3. Disconnect 4. Disadvantage 5. Disappoint 6. Disappear 7. Disown Etc.

Words repeated in the poem: 1. Strike 2. Give me the strength

RUSTUM AND SOHRAB Ans 1) Rustum was not able to live with his wife for a long time because he was called back by the king of Persia who didn’t feel safe unless Rustum was in his kingdom to protect it.

Ans 2) Tanimeh feared that Rustum would take the child away with him if he came to know that it was a boy. Tanimeh didn’t want to lose her child so she sent word to Rustum that their child was a daughter.

Ans 3) Sohrab learnt from his mother that he was the son of great Rustum, the Shield of Persia.

Ans 4) Tanimeh wanted Sohrab to show his father the precious stone tied on his arm. Sohrab’s father had given that stone to Tanimeh at the time of their parting.

Ans 5) Rustum had become old and had an aged father to look after. He also thought it improper to fight a beardless youth due to which he was reluctant to fight.

Ans 6) Rustum did not tell Sohrab who he was because he thought that if he told the truth Sohrab might want to make peace with him and withdraw from the fight.

Ans 7 ) Sohrab lost Courage when he heard Rustum’s war-cry. He was so unnerved that his shield fell off his hands and Rustum’s sword pierced his body.

Ans 8 ) Sohrab wished to be buried near his father’s home so that all the passers-by would say that the mighty Rustum’s son lay there whom his father had killed in ignorance.

B) One word substitution. i. Combat ii. Protect iii. Parting iv. Send word v. Determined vi. Longing vii. Invader viii.Distant ix. Breakdown x. Precious xi. Chief xii.Bury xiii. Mighty C.) Choose the correct option. (Answer key)

1. The king of Persia 2. The challenger 3. Tanimeh 4. Rustum 5.Sohrab D. Match the following ( answer key)

a. — viii b. — ix c. — x d. — vii e. — I f. — iii g. —ii h. — v i. — iv j. — vi Q) Combine the following pairs of sentences with unless.

Ans 1) Unless you run fast, you cannot catch the train. Ans 2) Unless you work hard, you cannot get a first class. Ans 3) Unless you hurry, you cannot catch the train. Ans 4) Unless you do as I tell you, you will regret it. Ans 5) Unless you tell me about your problem, I can’t give you some solution.


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Q) Match the columns. 1- D 2— f 3— a 4— b 5— c 6— e

Q) Report the following sentences. (answers)

1. She said her that her father would return from Jammu the next day. 2. They said that they would die for the sake of their country. 3. The teacher said that Babar won the first battle of Panipat. 4. I told him that he had made a false statement. 5. They told you that they would play a match the next day. 6. I told him that I was an early riser. 7. He said that his father had died the previous year. 8. She told me that the climate of that place didn’t suit her. 9. I told the peon that all his faults would be pardoned if he confessed them. 10. I said that I should finish my work as early as I could Q) Change the following sentences into direct speech. ( answers)

1. The employer said to him, “ You will be dismissed if you do not attend the office.” 2. Sanjay said, “ My brother has met with an accident yesterday.” 3. I said to him, “ I may not come tomorrow.” 4. The principal said, “ Tomorrow will be a holiday.” 5. The teacher said to us, “ You are intelligent and hardworking.”

GLOBAL WARMING- A CATASTROPHE IN THE MAKING

Ans 1) The Earth’s average surface temperature has increased due to the increasing concentration of greenhouse gases caused due to activities like deforestation and burning of fossil fuels.

Ans 2) The Himalayan region is called the water tower of Asia because it has the largest glacial coverage of about 33,000 sq. Kms., which provides about 8.6 cubic metres of water annually. It supports economy and livelihood of millions of people in several countries of Asia.

Ans 3) Flood plains, Lakes, high altitude Himalayan lakes, perennial rivers of Indus River water system, wetlands of different dimensions, ponds, snow-fed streams, springs, reservoirs are the main source of water in the Kashmir valley.

Ans 4) Water is the most precious natural resource because all biotic things depend on water for their survival. It helps in the economy, development, livelihood and is used to generate electricity. It is used for agriculture and drinking purposes. Without water no life is possible on earth.

Ans 5) Kashmir has been badly influenced by global warming. Water is becoming scarce due to the melting of glaciers such as Kolahoi glacier, the main source of water in Kashmir. Floods have increased and heavy wind storms are caused by global warming which destroys crops, fruits and affects the livelihood of people. Wetlands that were part of water for irrigation in Kashmir, have been concretised, thereby reducing the production of crops.

Ans 6) Global warming is the largest threat to humanity and also to all living organisms. Due to global warming, glaciers are melting, causing the rise in the sea level and the change in the precipitation in air.


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This will cause massive climatic changes throughout the world. The demarcation between seasons a decade ago has vanished. It is now difficult to distinguish between Spring and Summer, as also between Autumn and Winter. This will affect the crop yield which will cause food scarcity, starvation, diseases etc. Polar icecaps will melt and will affect coastal population.

Ans 7) Global warming leads to change in weather and climate. Snowfall is usually thin, late and does not last long. We can therefore say that global warming is the cause of decrease in snow fall in Kashmir.

Ans 8) Global warming has catastrophic effects on climate in J&K. Migratory birds have changed their cycle. Major lakes have shrunk and many small water bodies have completely disappeared. It has also affected the crop yield in J&K.

Ans 9) Scientists have considered the Kolahoi glacier as one of the index glaciers in Lidder Valley. It has been selected for long term monitoring. This glacier is the water resource of whole community in this valley. Scientific studies conducted on the glacier will last for five years after which TERI ( The Energy and Resources Institute in J&K) will recommend measures to bring down the glacial recession rate.

Ans 10) As individuals we should try to create awareness among the people about the catastrophic effects of global warming. Deforestation should be stopped, burning of fossil fuels minimised. All processes which are responsible for increase of Carbon Dioxide concentration should be minimised.

LANGUAGE WORK: Q) Fill in the blanks with compound words:

(Answers) 1.a) Green House b) greenhouse 2. a) white-collar b)white collar 3.a) white house b) White House 4.a) black box b) blackbox

5.a) strong hold b) stronghold Q) Use the phrases in your sentences.

1. In progress: The construction of the new building is in progress. 2. Uncontrollable rate: The price of things is increasing at an uncontrollable rate. 3. Across the globe: The impact of global warming is visible across the globe. 4. In the backdrop of: In the backdrop of looming water crisis, we can’t afford to waste water. 5. At an alarming rate: The glaciers are receding at an alarming rate. 6. An irritating change: Pollution leads to an irritating change in the environment. 7. Bereft of: The mountains are bereft of any snow 8. Dried up: Most of the springs in the valley have dried up. 9. Heavy burden: The children will bear the heavy burden of our indifference. 10. At the global level: Steps for controlling pollution should be taken at the global level. Q) Choose the synonyms. (answers)

1. Irregular 2. Suited 3. Repetitive 4. Quay Q) Use infinitive or gerund.

1. Laughing 2. Being 3. To go 4. Being 5. Having 6. Listening 7. To read 8. To cheat 9. Fishing 10. To make


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SUBJECT – MATHEMATICS Squares and Square Roots EXERCISE 6.1 1. What will be the unit digit of the squares of the following numbers? (i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387

(vii) 52698 (viii) 99880 (ix) 12796 (x) 55555 Sol: (i) The unit’s digit of (81)2 will be 1. (ii) The unit’s digits of (272)2 will be 4. (iii) The unit’s digit of (799)2 will be 1. (iv) The unit’s digit of (3853)2 will be 9. (v) The unit’s digit of (1234)2 will be 6. (vi) The unit’s digit of (26387)2 will be 9. (vii) The unit’s digit of (52698)2 will be 4. (viii) The unit’s digit of (99880)2 will be O. (ix) The unit’s digit of (12796)2 will be 6. (x) The unit’s digit of (55555)2 will be 5. 2. The following numbers are obviously not perfect squares. Give reason. (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000

(vi) 89722 (vii) 222000 (viii) 505050 Sol: (i) 1057 Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9) ∴1057 is not a perfect square. (ii) 23453 ------------ Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9). ∴23453 is not a perfect square. (iii) 7928 -------------- Since, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9). ∴7928 is not a perfect square. (iv) 222222 -----------------Since, the ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9). ∴222222 is not a perfect square. (v) 64000 ------------------------ Since, the number of zeros is odd. ∴64000 is not a perfect square. (vi) 89722 -----------------Since, the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9). ∴89722 is not a perfect square. (viii) 222000 ----------------Since, the number of zeros is odd. ∴222000 is not a perfect square. (viii) 505050 The unit’s digit is odd zero. ∴505050 cannot be a perfect square.


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3. The squares of which of the following would be odd numbers? (i) 431 (ii) 2826 (iii) 7779 (iv) 82004 Sol: Since the square of an odd natural number is odd and that of an even number is an even number. (i) The square of 431 is an odd number. [∵ 431 is an odd number.]

(ii) The square of 2826 is an even number. [∵ 2826 is an even number.]

(iii) The square of 7779 is an odd number. [∵ 7779 is an odd number.]

(v) The square of 82004 is an even number. [∵ 82004 is an even number.] 4. Observe the following pattern and find the missing digits. Sol: Observing the pattern, we have (i) (100001)2 = 10000200001 (ii) (10000001)2 = 100000020000001 5. Observe the, following pattern and supply the missing number. Sol: Observing the patter, we have (i) (1010101)2 = 1020304030201 (ii) 10203040504030201 = (101010101)2

6. Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32 22 + 32 + 62 = 72 32 + 42 + 122 =132 42 + 52 + —2 = 212 52 + —2 + 302 = 312 62 + 72 + —2 = —2 Note: To find pattern: Third number is related to first and second number. How? Fourth number is related to third number. How? Sol: The missing numbers are (i) 42 + 52 + 202 = 212 (ii) 52 + 22 + 302 = 312 (iii) 62 + 72 + 422 = 432 7. Without adding, find the sum. Sol: (i) 1 + 3 + 5 + 7 + 9 = 52 = 25 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 102 = 100 (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 + 21 + 23 = 122 = 144 8. (i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers. Sol: (i) 49 = 72 = Sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 (ii) 121 = 112 = Sum of first 11 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21


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9. How many numbers lie between squares of the following numbers? (i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100 Sol: Since between n2 and (n + 1)2, there are 2n non-square numbers. (i) Between 122 and 132, there are 2 × 12, i.e. 24 numbers (ii) Between 252 and 262, there are 2 × 25, i.e. 50 numbers (iii) Between 992 and 1002, there are 2 × 99, i.e. 198 numbers EXERCISE 6.2 1. Find the square of the following numbers. (i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46 Sol: (i) (32)2 = 32 x 32 = 1024 (ii) (35)2 = 35 x 35 = 1225 Do rest of the parts. 2. Write a Pythagorean triplet whose one member is (i) 6 (ii) 14 (iii) 16 (iv) 18 Sol: (i) Let 2n =6 ∴n=3 Now, n2 – 1 = 32 –1 = 8 and n2 + 1 = 32 + 1 = 10 Thus, the required Pythagorean triplet is 6, 8, 10. (ii) Let 2n = 14 ∴n = 7 Now, n2 – 1 = 72 – 1 = 48 and n2 + 1 = 72 + 1 = 50 Thus, the required Pythagorean triplet is 14, 48, 50. (iii) Let 2n = 16 ∴ n = 8 Now, n2 – 1 = 82 – 1 = 64 – 1 = 63 and n2 + 1 = 82 + 1 =64 + 1 = 65 ∴ The required Pythagorean triplet is 16, 63, 65. (iv) Let 2n = 18 ∴ n = 9 Now, n2 – 1 = 92 – 1 =81 – 1 = 80 and n2 + 1 = 92 + 1 = 81 + 1 = 82 ∴ The required Pythagorean triple is 18, 80, 82. EXERCISE 6.3 1. What could be the possible ‘one’s digits of the square root of each of the following numbers? (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025 Sol: The possible digit at one’s place of the square root of: (i) 9801 can be 1 or 9. …….. [∵ 1 × 1 = 1 and 9 × 9 = 81]


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(ii) 99856 can be 4 or 6. ………… [∵ 4 × 4 = 16 and 6 × 6 = 36] (iii) 998001 can be 1 or 9. (iv) 657666025 can be 5. [∵ 5 × 5 = 25] 2. Without doing any calculation, find the numbers which are surely not perfect squares. (i) 152 (ii) 257 (iii) 408 (iv) 441 Sol: We know that the ending digit of perfect square is 0, 1, 4, 5, 6, and 9. ∴ A number ending in 2, 3, 7 or 8 can never be a perfect square. (i) 153, cannot be a perfect square. (ii) 257, cannot be a perfect square. (iii) 408, cannot be a perfect square. (iv) 441, can be a perfect square. 3. Find the square roots of. 100 and 169 by the method of repeated subtraction. Sol: (i) 100 -1 = 99 – 3 = 96 – 5 = 91 – 7 = 83 – 11 = 72 – 13 = 59 – 15 = 45 – 17 = 38 – 17 = 19 – 19 = 0 ∴ 100 = (10)2 (ii) 169 – 1 = 168 – 3 = 165 – 5 = 160 – 7 = 153 – 9 = 144 – 11 = 133 – 13 = 120 – 15 = 105 – 17 = 88 – 19 = 69 – 21 = 48 – 23 = 25 – 25 = 0 ∴ 169 = (13)2 4. Find the square roots of the following numbers by the Prime Factorisation Method. (i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100 Sol: (i) 729 = √3 x 3 x 3 x 3 x 3 x 3 = √3 x 3 x 3 = 3 x 3 x 3 = 27 (ii)

400 = √2 x 2 x 2 x 2 x 5 x 5 = √2 x 2 x 5 = 2 x 2 x 5 = 20 Do rest of the parts yourself.

5. For each of the Following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained. (i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768

3 729 3 243 3 81 3 27 3 9 3 3 1

2 400 2 200 2 100 2 50 5 25 5 5 1


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Sol: (i) We have 252 = 2 × 2 × 3 × 3 × 7 ∵ The prime factor 7 has no pair. ∴ [252] × 7= [2 × 2 × 3 × 3 × 7] × 7 or 1764 = 2 × 2 × 3 × 3 × 7 × 7 Thus, the required smallest whole number = 7 Also, the square root of 1764 is 42 Do rest of the parts yourself. 6. For each of the following numbers, find the smallest whole number by which it should be divided so

as to get a perfect square. Also find the square root of the square number so obtained. (i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620 Sol: (i) 252 We have 252 = 2 × 2 × 3 × 3 × 7 ∵ The prime factor 7 has no pair. ∴ [252] × 7= [2 × 2 × 3 × 3 × 7] ÷ 7 or 36 = 2 × 2 × 3 × 3 Thus, the required smallest whole number = 7 Also, the square root of 36 is 6. Do rest of the parts yourself.

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Sol: Let the number of students = x Each student donated Rs x. Total amount donated by the class = Rs x × x = Rs x2

2 252 2 126 3 63 3 21 7 7 1

2 252 2 126 3 63 3 21 7 7 1


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Thus, x2 = 2401 Or x = 7 x 7 = 49 Hence there are 49 students in the class. 8. 2025 plants are to be planted in a garden in such a tivav that each row contains as many plants as the

number of rows. Find the number of row and the number of plants in each row. Sol: Let the number of rows = x ∴ Number of plants is a row = x So, the number of plants to be planted = x × x = x2 i.e., x2 = 2025 Or x = 3 x 3 x 5 = 45 Thus, the required number of rows = 45 Also, number of plants in a row = 45. 9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10. Sol: We know that LCM is the smallest number divisible by all its factors. Since, LCM of 4, 9 and 10 = 2 × 2 × 9 × 5 = 180 But 180 is not a perfect square. Again, ∴ 180 = 2 × 2 × 3 × 3 × 5 To make it a perfect square we will multiply 180 by 5 i.e., 180 x 5 = 900 ∵ All the prime factors of 900 are paired. ∴ 900 is a perfect square. Thus, the required number = 900. 10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20 Sol: The smallest number divisible by 8, 15 and 20 is their LCM. We have LCM = 2 × 2 × 5 × 2 × 3 = 120 But 120 is not a square number. Now, to make it a perfect square, we have 120 = 2 × 2 × 2 × 3 × 5 or [120] × 2 × 3 × 5 = [2 × 2 × 2 × 3 × 5] × 2 × 3 × 5 or 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 All factors of 3600 are paired. Therefore, 3600 is a perfect squared. ∴ The required number = 3600.

7 2401 7 343 7 49 7 7 1

3 2025 3 675 3 225 3 75 5 25 5 5 1


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EXERCISE 6.4

1. Find the square root of each of the following numbers by Division method. (i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369 (vii) 5 776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900

Do rest of the parts yourself 2. Find the number of digits in the square root of each of the. following numbers (without any calculation). (i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625 Sol: If ‘n’ stands for number of digits in the given number, then

3. Find the square root of the following decimal numbers. (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36


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Do rest of the parts yourself . 4. Find the least nti,nllc’r” which must be subtracted from each of the following numbers. so as to get a

perfect square. Also find the square root of the perfect square so obtained. (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000 Sol: (i) On proceeding to find the square root of 402, we have Since, we get a remainder 2 The required least number to be subtracted from 402 is 2. Do rest of the parts yourself. 5. Find the least number which must be added to each of the following numbers so as to get a perfect

square. Also find the square root of the perfect ware so obtained. (i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412

Do rest of the parts yourself.


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6. Find the length of the side of a square whose area is 441 m2. Sol: Let the side of the square = x metre Area = side × side

441 = x x x

7. In a right triangle ABC, ∠B = 90°. (a) If AB = 6 cm, BC = 8 cm, find AC.

(b) If AC = 13 cm, BC = 5 cm, find AB. Sol:

8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more or this.


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Sol: Since, the number of plants in a row and the number of columns are the same.

9. There are 500 children in a school. For a P T drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children this arrangement? Sol: Since, the number of rows and the number of columns are same.

Total number (i.e. their product) must be a square number, we have Since, we get a remainder of 16 500 > (22)2 or 500 – 16 = (22)2 Thus, the required number of children to be left out = 16


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Cubes and Cube Roots Exercise 7.1

Q1. Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656 Sol. (i) We have 216 = 2 × 2 × 2 × 3 × 3 × 3 Grouping the prime factors of 216 into triples, no factor is left over. ∴ 216 is a perfect cube. (ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor. ∴ 128 is not a perfect cube. (iii) We have 1000 = 2 × 2 × 2 × 5 × 5 × 5 Grouping the prime factors of 1000 into triples, we are not left over with any factor. ∴ 1000 is a perfect cube. (iv) We have 100 = 2 × 2 × 5 × 5 Grouping the prime factors into triples, we do not get any triples. Factors 2 × 2 and 5 × 5 are not in triples. ∴ 100 is not a perfect cube. (v) We have 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 Grouping the prime factors of 46656 in triples we are not left over with any prime factor. ∴ 46656 is a perfect cube. Q2. Find the smallest number by which each of the following numbers must be multiplied to obtain a

perfect cube. (i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100 Sol. (i) We have 243 = 3 × 3 × 3 × 3 × 3 The prime factor 3 is not a group of three. ∴ 243 is not a perfect cube. Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3 or 729 =3 × 3 × 3 × 3 × 3 × 3 Now, 729 becomes a perfect cube. Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 Grouping the prime factors of 256 in triples, we are left over with 2 × 2. ∴ 256 is not a perfect cube. Now, [256] × 2 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2 or 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 i.e. 512 is a perfect cube. Thus, the required smallest number is 2.

(iii) We have 72 = 2 × 2 × 2 × 3 × 3 Grouping the prime factors of 72 in triples, we are left over with 3 × 3. ∴ 72 is not a perfect cube. Now, [72] × 3 = [2 × 2 × 2 × 3 × 3] × 3 or 216 = 2 × 2 × 2 × 3 × 3 × 3


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i.e. 216 is a perfect cube. ∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

(iv) We have 675 = 3 × 3 × 3 × 5 × 5 Grouping the prime factors of 675 to triples, we are left over with 5 × 5. ∴ 675 is not a perfect cube. Now, [675] × 5 = [3 × 3 × 3 × 5 × 5] × 5 or 3375 = 3 × 3 × 3 × 5 × 5 × 5 Now, 3375 is a perfect cube. Thus, the smallest required number to multiply 675 such that the new number perfect cube is 5.

(v) We have 100 = 2 × 2 × 5 × 5 The prime factor are not in the groups of triples. ∴100 is not a perfect cube. Now [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5 or [100] × 10 = 2 × 2 × 2 × 5 × 5 × 5 1000 = 2 × 2 × 2 × 5 × 5 × 5 Now, 1000 is a perfect cube. Thus, the required smallest number is 10. Q3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect

cube. (i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704 Sol. (i) We have 81 = 3 × 3 × 3 × 3 Grouping the prime factors of 81 into triples, we are left with 3. ∴ 81 is not a perfect cube. Now, [81] 3 = [3 × 3 × 3 × 3] + 3 or 27 = 3 × 3 × 3 i.e. 27 is a perfect cube Thus, the required smallest number is 3. (ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 Grouping the prime factors of 128 into triples, we are left with 2. ∴ 128 is not a perfect cube Now, [128] 2 = [2 × 2 × 2 × 2 × 2 × 2 or 64 = 2 × 2 × 2 × 2 × 2 × 2 i.e. 64 is a perfect cube. ∴ The smallest required number is 2. (iii) We have 135 = 3 × 3 × 3 × 5 Grouping the prime factors of 135 into triples, we are left over with 5. ∴ 135 is not a perfect cube Now, [l35] 5 = [3 × 3 × 3 × 5] 5 or 27 = 3 × 3 × 3 i.e. 27 is a perfect cube. Thus, the required smallest number is 5. (iv) We have 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 Grouping the prime factors of 192 into triples, 3 is left over. ∴ 192 is not a perfect cube.


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Now, [192] 3 =[2 × 2 × 2 × 2 × 2 × 2 or 64 = 2 × 2 × 2 × 2 × 2 × 2 i.e. 64 is a perfect cube. Thus, the required smallest number is 3. (v) We have 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 Grouping the prime factors of 704 into triples, 11 is left over. ∴ [704] 11 =[2 × 2 × 2 × 2 × 2 × 2 or 64 = 2 × 2 × 2 × 2 × 2 × 2 i.e. 64 is a perfect cube. Thus, the required smallest number is 11. Q4. Parikshit makes a cuboid of plasticine of sidec 5 cm, 2 cm, 5 cm. How many such cuboids will he

need to form a cube? Sol: Sides of the cuboid arc: 5 cm, 2 cm, 5 cm ∴ Volume of the cuboid = 5 cm × 2 cm × 5 cm To form it as a cube its dimension should be in the group of triples. ∴ Volume of the required cube = [5 cm × 5 cm × 2 cm] × 5 cm × 2 cm × 2 cm =[5 × 5 × 2 cm3] = 20 cm3 Thus, the required number of cuboids = 20. Exercise 7.2 1. Find the cube root of each of the following numbers by prime factorisation method. (i) 64 (ii) 512 (iii) 10648 (iv) 27000 (v) 15625 (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125 Sol. (i) 64 By prime factorisation, we have √64 = √2 x 2 x 2 x2 x2 x 2 = √2 x 2 = 2 x 2 = 4 (iv) 27000 By prime factorisation, we have √27000 = √2 x 2 x 2 x 3 x 3 x 3 = √2 x 3 x 5 = 2 x 3 x 5 = 30 Do rest of the parts yourself . Q2. State true or false. (i) Cube of any odd number is even. (ii) A perfect cube does not end with two zeros. (iii) If square of a number ends with 5, then its cube ends with 25. (iv) There is no perfect cube which ends with 8.

2 64 2 32 2 16 2 8 2 4 2 2 1

2 27000 2 13500 2 6750 3 3375 3 1125 3 375 5 125 5 25 5 5 1


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(v) The cube of a two digit number may he a three digit number. (vi) The cube of a two digit number may have seven or more digits. (vii) The cube of a single digit number may be a single digit number. Sol. (i) False (ii) True (iii) False (iv) False (v) False (vi) False (vii) True Q3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root?

Similarly, guess the cube roots of 4913, 12167, 32768. Sol. (i) Separating the given number (1331) into two groups: 1331 → 1 and 331 ∴ 331 end in 1. ∴ Unit's digit of the cube root = 1 ∴ Ten's digit of the cube root = I (ii) Separating the given number (4913) in two groups: 4913 → 4 and 913 Unit's digits: ∵ Unit's digit in 913 is 3. ∴ Unit's digit of the cube root = 7 [73 = 343; which ends in 3] Ten's digit: 13 = 1, 23 = 8 and 1 < 4 < 8 i.e. 13 < 4 < 23 ∴ The ten's digit of the cube root is 1. (iii) Separating 12,167 in two groups: 12167 → 12 and 167 Unit's digit: ∵ 167 is ending in 7 and cube of a number ending in 3 ends in 7. ∴ The unit's digit of the cube root = 3 Ten's digit: ∵ 2

3 = 8 and 33 = 27 Also, 8 < 12 < 27 or 23 < 12 < 32 ∴ The tens digit of the cube root can be 2. (iv) Separating 32768 in two groups: 32768 → 32 and 786 Unit's digit: 768 will guess the unit's digit in the cube root. ∵ 768 ends in 8. Unit's digit in the cube root = 2 Ten's digit: ∵ 3

3 = 27 and 43 – 64 Also, 27 < 32 < 64 or 33 < 32 < 43 The ten's digit of the cube root = 3.


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Comparing Quantities EXERCISE 8.1

1. Find the ratio of the following: (a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour. (b) 5 m to 10 km (c) 50 poise to Rs 5 Solution: In a ratio, the quantities are in the same unit if they are not in the same units, then first we convert them

in the same unit.

2. Convert the following ratios to percentages. (a) 3 : 4 (b) 2 : 3 Solution:

3. 72% of 25 students are good in Mathematics. How many are not good in Mathematics? Solution:


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4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Solution:

5. If Insha had Rs 600 left after spending 75% of her money, how much did she in the beginning?

Solution: Insha made spending of Rs 75%. She is left with Rs (100 – 75)% or Rs 25%. But she is having Rs 600 now. .

6. If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.

Solution: People who like cricket = 60% People who like football = 30%


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Exercise 8.3


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i.e., n = 2 x = 3

Do rest of the parts yourself.


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SUBJECT – SCIENCE Reaching the age of Adolescence Q#1 What is the term used for secretions of endocrine glands responsible for changes taking place in the

body? Ans. Puberty Q#2 Define adolescence. Ans. The period of life, when the body undergoes changes, leading to reproductive maturity is called

adolescence. Q#3 What is menstruation? Explain. Ans. Adolescents become capable of reproduction when their testes and ovaries begin to produce

gametes. The capacity for maturation and production of gametes lasts for a much longer time in males than in females. The ova begins to mature with the on-set of puberty. One ovum matures and is released by one of the ovaries once in about 28 to 30 days. During this period, the wall of the uterus becomes thick so as to receive the egg, in case it is fertilized and begins to develop. This is known as pregnancy. If fertilization does not occur, the unfertilized egg, and the thickened lining of the uterus along with its blood vessels are shed off. This causes bleeding in women which is called menstruation.

Q#4 List changes in the body that take place at puberty. Ans. The changes in the body that take place at puberty are:

i) Increase in height.

ii) Change in body shape.

iii) Change in voice.

iv) Increased activity of sweat and sebaceous glands.

v) Development of sex organs.

vi) Production of hormones.

vii) Reaching mental, intellectual and emotional maturity.

Q#5 What are sex hormones? Why are the named so? State their function. Ans. Sex hormones are the hormones that are produced by sex organs i.e. Testes and ovaries.

The testes produce the male sex hormone called testosterone and the ovaries produce the female sex hormone called estrogen.

Function of testosterone is to promote sperm production and help in the development of secondary sexual characters and the function of ovaries is to promote egg formation and help in the development of secondary sexual characters.


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Q#6 Choose the correct option.

a) Adolescents should be careful about what they eat, because i) Proper diet develops their brains. ii) Proper diet is needed for the rapid growth taking place in their body. iii) Adolescents feel hungry all the time. iv) Taste buds are well developed in teenagers.

Ans. ii) Proper diet is needed for the rapid growth taking place in their body.

b) Reproductive age in women starts when their i) Menstruation starts. ii) Breasts start developing. iii) Body weight increases. iv) Height increases.

Ans. i) Menstruation starts.

c) The right meal for adolescents consists of i) Chips, noodles, coke.

ii) Chapatti, dal, vegetables.

iii) Rice, noodles and burger.

iv) Vegetable cutlets, chips and lemon drink.

Ans. ii) Chapatti, dal, vegetables. Q#7 Write notes on:

a) Adam’s apple

b) Secondary sexual characters

c) Sex determination in the unborn baby.

Ans. a) Adam’s Apple:- This is the protruding part of the throat in boys which begins to grow at puberty. This is also known as larynx or voice box.

d) Sex determination in the unborn baby: - All human beings have 23 pairs of chromosomes as

the name X and Y, female has two X chromosomes and male has one X and one Y chromosome. When a sperm containing X chromosome fertilizes the egg, the zygote will have two X chromosomes and will develop into a female child. If the sperm contributes a Y chromosome to the egg (ovum) at fertilization, the zygote will develop into a male child.


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METALS & THEIR PROPERTIES “Additional Questions”

Physical Properties of Metals:- i) Physical State:- Metals are generally solid at room temperature except Mercury and Gallium

which are a liquid at room temperature. ii) Hardness: - Most metals are very hard. However Na & K are exception both are so soft that

they can be easily cut with a knife. iii) Opaque nature:- Generally metals are opaque in nature & do not allow the light to pass

through them. iv) Metallic Lusture: - The freshly cut & uncorroded surface of metals is lustrous as they are

good reflectors. v) Density: - Metals have relatively high densities except Sodium & Potassium. vi) Melting Point & Boiling Point: - Metals have high melting point & boiling point except Na

& K both melt & boil at low temperature. vii) Thermal & electrical conductivity. All metals are good conductors of heat & electricity. It is

found that Ag is best conductor & lead (Pb) is poorest. The metals are good conductors because they contain free or mobile electrons which can conduct heat & electricity.

viii) Malleability :- Metals are malleable i.e. they can be beaten into very thin sheets called foils with a hammer without breaking Al & CU are highly malleable.

ix) Ductility: - Metals are ductile i.e./ they can be drawn into thin wires Silver & Gold are best ductile.

x) Metals are sonorous:-i.e. they produce a ringing sound when struck hard on surface of solid. xi) Tensile strength: - metals have high tensile strength. Due to this property of metals, heavy

weights can be transferred into metallic wires without breaking. Chemical Properties Of Metals:- The metals are electropositive in nature. The tendency of metal atoms to form positive ions gives rise to characteristic chemical properties of metals.

e.g. 11

Na Na+ + 1 e- (2,8,1) K,L,M 12 Mg Mg2+ + 2e/ (2,8,2) 13 Al Al 3+ + 3e/ (2,8,3)

1)Reaction with Oxygen:- Almost all metals combine with oxygen to form metal oxides. The general

equation is Metal + Oxygen Metaloxide.

All metals do not react with same speed.

1) K & Na reacts so vigorously that they catch fire if kept in open air.


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4K(s) +O2 (g) 2K2O(s)

4Na(s) + O2 (g) 2Na2 O (s)

2) When Mg is heated it burns with a blinding white light to form Magnesium Oxide.

2 Mg(s) +O2 (g) 2MgO

3) Al combines with O2 only on heating

4Al + 3O2 2Al2O3

4) Zn burns in air only on strong heating.

2Zn(s) + O2 (g) 2ZnO

5) Iron does not burn on heating but glows brightly 3Fe + 2O2 Fe3O4 6) Copper does not burn , but a layer of black substance known as copper oxide are formed. 2Cu(s) + O2 2CuO

7) Silver & Gold do not react with O2 even at high temperature.

2) Reaction with Water:- Metals react with water to form Metal Oxide or Metal Hydroxide with

the liberation of Hydrogen gas.

Metal + Water Metal hydroxide + Hydrogen

i) Na metals react vigorously with cold H2O

2Na (s) + 2H2O (l) 2NaOH + H2 (g) Sodium Hydroxide.

b) Magnesium metals reacts rapidly with hot boiling water Steam

Mg(s) + 2H2O (g) Mg (OH)2 + H2 Magnesium hydroxide Mg(s) + 2H2O (g) Hot boiling water MgO + H2 Magnesium Oxide

c) Zn metal reacts only with steam Steam

Zn + H2O (g) ZnO + H2 Zinc Oxide


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Iron metal does not react with water under ordinary conditions. The reaction occurs only when steam is passed over red hot iron. 3Fe + 4H2O Fe3O4 + 4H2 (g)

(Fe2O3+ FeO) Copper metal does not react with water even under strong conditions.

3) Reaction with Acids When a metal reacts with dilute mineral acid metal salt is formed & hydrogen gas is liberated.

Metal + Acid Metal salt + H2.

i) Na reacts violently with dilute HCl or dilutes H2SO4 to form metal salt & H2.

2Na + 2HCl 2NaCl +H2 2Na + H2SO4 Na2SO4 +H2

ii) Mg reacts quite rapidly with HCl & H2SO4

Mg + 2HCl MgCl2 + H2 Mg + H2SO4 MgSO4 + H2

iii) Cu does not react with acid like Ag &Au.

Q#1 State two physical properties on the basis of which metals may be distinguished from non-metals.

Metals Non-Metals i. Metals are good conductors of

electricity. ii. Metals are generally solids at room

temperature except mercury and gallium which is liquid at room temperature.

i. Non-metals are bad conductors of electricity except graphite (a form of carbon) which is a good conductor of electricity.

ii. Non-metals occur as solids (carbon, sulphur, Iodine), liquids (bromine) or gases (oxygen, nitrogen).

Q#2 State one chemical property which may be used to distinguish a metal from a non-metal. Ans. One chemical property on the basis of which we can distinguish a metal from a non-metal is as

follows: Whenever metals react with oxygen they form basic oxide (metallic oxide) e.g. 2MgO + O2 → 2MgO (basic oxide) Magnesium Oxygen Magnesium oxide 4Na + O2 → 2Na2O Sodium Oxygen Sodium oxide 2Zn + O2 → 2ZnO


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Zinc Oxygen Zinc Oxide Whenever non-metals react with oxygen they form acidic oxide or neutral oxide (non-metallic

oxide) e.g. C + O2 → CO2 (acidic oxide) Carbon Oxygen Carbon dioxide S + O2 → SO2(acidic oxide) Sulphur Oxygen Sulphur dioxide 4P + 5 O2 → 2P2O5(acidic oxide) Phosphorus Oxygen Phosphorus pentoxide 2H2 + O2 → 2H2O(neutral oxide) Q#3 Why is aluminium used for making cooking utensils? Ans. Aluminium being a metal is a good conductor of heat as it has free electrons which help it to

conduct heat. That is why aluminium is used for making cooking utensils. Q#4 What are noble metals? Is silver a noble metal? Ans. Those metals which occur in native or free state in nature and don’t react with air, water and acids

are called noble metals. Yes, silver is a noble metal. Q#5 Name a non-metal which is a good conductor of electricity. Ans. Graphite is a non-metal which is good conductor of electricity. Q#6 Name two non-metals which occur in a solid state at room temperature. Ans. Sulphur and carbon are two non-metals which occur in a solid state at room temperature. Q#7 What is the gas evolved when a metal reacts with acid? Ans. When a metal reacts with acid, it forms corresponding salts along with the liberation of hydrogen

gas. 2Al + 6HCl → 2AlCl3 + 3H2

Mg + HCl → MgCl2 + H2 Q#8 Magnesium and copper metals are directly heated over a flame which of these will burn in air.

Which is more reactive? Ans. Magnesium will burn in air as it is more reactive. Q#9 Why is gold found in the native state in the earth’s crust, but not iron? Ans. Gold is a noble metal so it possesses a characteristic of being chemically inert i.e. it doesn’t react

with air, water and acids. That is why it is found in native (free) state in nature while as iron reacts with air, water and acids to form compounds. Therefore, it is not found in native state in nature.

Q#10 Why is sodium stored in kerosene oil? Ans. Sodium is the most reactive metals, so it spontaneously even at room temperature reacts with

oxygen to form sodium oxide (Na2O). To prevent this reaction of sodium it is stored in kerosene oil.


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Q#11 Describe how magnesium reacts with dilute hydrochloric acid. Write the equation for this reaction. Ans. Magnesium is a reactive metal so it vigorously reacts with hydrochloric acid giving magnesium

chloride (a salt) and hydrogen gas Mg + 2HCl → MgCl2 + H2 Q#12 How would you proceed to prepare hydrogen in the laboratory? Ans. In laboratory, hydrogen gas is prepared by the reaction of zinc with dilute hydrochloric acid (HCl) . Zn + 2HCl → ZnCl2 + H2 Place a few pieces of zinc in a bottle having a wide mouth. Fit a two bored cork. Insert a thistle

funnel from one bore and a delivery tube from the other, as shown in figure. Now, pour dilute HCl in the bottle through thistle funnel and collect the hydrogen gas evolved by the downward displacement of water.

Q#13 What would you expect to observe when a strip of zinc is dipped in a copper sulphate solution? Ans. Zinc is more reactive than copper. Thus, when a strip of zinc is dipped in a copper sulphate

solution, a brown coating of copper is formed over the strip and blue copper sulphate solution turns colourless.

Q#14 Why is iron not deposited over a copper plate when the latter is dipped in a ferrous sulphate

solution? Ans. Iron is more reactive metal than copper. More reactive metals replace less reactive metals from

their salt solution but less reactive metals cannot replace more reactive metals. So iron is not deposited over a copper plate when copper plate is dipped in a ferrous sulphate solution.

Q#15 Arrange the following metals in decreasing order of their reactivity. Which of these is the least

reactive? a. Iron b. Sodium c. Copper d. Magnesium Ans. Decreasing order: Sodium, magnesium, iron, copper. Copper is the least reactive metal. Q#16 Why does an aluminium vessel lose its shine so soon after use? Ans. Aluminium being a reactive metal begins to corrode as soon as it comes in contact with moist air. It

combines with moist air to form a coating of aluminium oxide. 4Al + H2O 2Al2O3 + 6H2

This layer of aluminium oxide covers the upper surface of the aluminium so it loses its shine soon after its use.

Q#17 Silver does not combine easily with oxygen but silver jewellery tarnishes after some time. How? Ans. Silver does not combine easily with oxygen but it readily combines with sulphur compounds such

as hydrogen sulphide present in the air and forms a black coating of silver sulphide. That is why the jewellery made from it tarnishes after sometime.

2Ag + H2S Ag2S + H2 Q#18 Which one of the following materials is most likely to be corroded? Ans. a. a wood plank b. a steel chair. c. an exposed iron rod d. an iron rod coated with oil. Q#19 Name a common metal which is highly resistant to corrosion. Ans. Titanium, also called as strategic metal is highly resistant to corrosion. Q#20 What is corrosion? What are the ways to retard this process?


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Ans. Corrosion is defined as the gradual transformation of a metal into its combined state because of its reaction with environment.

Or The eating away of metals layer by layer by the action of air and moisture on their surface is called corrosion. There are several ways to retard this process [corrosion] i. Painting: Painting the surface of the metal after cleaning it thoroughly is the most common

method of preventing corrosion because the paint does not allow the metal to come in contact with the moist air.

ii. Applying grease or oil: Applying grease or oil has a similar effect as a coating of grease or oil also protects metals from corrosion by shutting out air & moisture.

iii. Galvanizing:- Another method of preventing iron from rusting is to galvanize it. It is the process in which highly reactive metal is coated over less reactive metal. It is mostly done in iron. A thin coating od zinc is coated over iron which prevents it from rusting.

iv. Tin Coating:- Tin sheets are dipped in molten zinc and are taken out and allowed to cool to get tin coated iron.

v. Electroplating:- It is the process by which metal surfaces are covered with layers of other metals {which are resistant to corrosion}by passing an electric current e.g. Iron and Steel are protected from corrosion by coating them with a layer of tin or chromium metal which are themselves resistant to corrosion.

vi. Alloying: Some metals become more corrosion resistant when alloyed (mixed) with other metals e.g. stainless steel which is an alloy of iron, chromium & nickel does not rust easily and brass which is alloy of copper and zinc does not also rust easily.

Q#21 Arrange the elements – copper, zinc, lead, calcium, potassium, silver & aluminium in order of their

increasing reactivity with oxygen and water. Write the reaction of one metal of choice with oxygen and water.

Ans. Silver, copper, lead, zinc, aluminium, calcium, potassium. Reaction of magnesium with oxygen and water:

i. 4K + O2 → 2K2O Potassium Oxygen Potassium Oxide ii. 2K + 2H2O → 2KOH + H2

Potassium Water Potassium Hydroxide Hydrogen Q#22 From amongst the set of metals sodium, zinc, iron, copper – Select the following giving equations

for each reaction: - a. Two metals which will liberate hydrogen from water. Ans. Sodium and Iron. Reactions: 2Na + 2H2O → 2NaOH + H2 2Fe + 3H2O → Fe2O3 + 3H2 b. One metal which is used to prepare hydrogen gas in the laboratory. Ans. Zinc is used to prepare hydrogen gas in the laboratory Zn + 2HCl → ZnCl2 + H2 c. One metal which will displace copper from copper sulphate solution. Ans. Zinc will displace copper from copper sulphate solution. Zn + CuSO4 → ZnSO4 + Cu


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d. One metal which will not displace copper from copper sulphate solution. Ans. Copper will not displace copper from copper sulphate solution. The other three metals are more

reactive than copper and thus, can replace it from its salt solution. Q#23 Suggest reasons why a. Silver is used in jewellery. Ans. Silver is a metal so it has a shiny appearance i.e. it shows metallic luster and does not easily react

with air, water and acids as other metals do. That is why it is used in jewellery. b. Copper is used for electrical wiring Ans. Copper is a metal and metals are good conductors of electricity. Therefore, it is used for electrical

wiring. c. Gold is found in the native state. Ans. Gold is found in the native state because it does not easily react with air, water and acids. d. Sodium does not occur in the free state in nature. Ans. Sodium does not occur in the free state in nature because it is a highly reactive metal and reacts

with air, water and acids thereby forming compounds such as chlorides, oxides and sulphides. Q#24 Give two uses of highly sonorous metals. Ans. The two uses of highly sonorous metals are: -

i. Their wires are used in musical instruments. ii. They are used for making bells.

Q#25 Describe how magnesium and iron react with oxygen. Write the equation for their reactions. Ans. Magnesium metal does not react with oxygen at room temperature. On heating, it readily burns in

oxygen with a bright white light producing magnesium oxide (MgO). 2Mg + O2 → 2MgO Iron doesn’t combine with oxygen at room temperature. When iron is strongly heated in oxygen, it

forms a brown black compound called iron oxide, Fe2O3. 4Fe + 3O2 → 2Fe2O3 Q#26 What would happen if iron powder is heated with copper oxide? Write a balanced equation of this

reaction. Ans. If iron powder is heated with copper oxide, copper is replaced to form iron oxide and copper. 2Fe + 3CuO → 3Cu + Fe2 O3 Q#27 Give an example which shows that magnesium is more reactive than copper. Ans. When copper oxide is heated with magnesium powder, magnesium oxide and copper are formed,

this show that magnesium is more reactive than copper. Mg + CuO → Mg O + Cu Q#28 Name one metal which will fit each of the following descriptions choosing a different metal each

time. Also, write the equation of reaction: a. A metal which floats on water reacts vigorously with it and forms an alkali. Ans. Sodium is a metal which floats on water, reacts vigorously with it and forms an alkali 2Na + 2H2O → 2NaOH + H2


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Sodium water Sodium hydroxide (alkali) hydrogen b. A metal that replaces silver from silver nitrate solution. Ans. Copper is a metal that replaces silver from silver nitrate solution Cu + 2AgNO3 → 2Ag + Cu (NO3)2 c. The metal which is used for galvanizing iron. Ans. Zinc is the metal which is used for galvanizing iron. Fe + Zn (molten) → galvanized iron d. A metal that reacts with oxygen without burning. Ans. Copper is a metal that reacts with oxygen without burning. 2Cu + O2 → 2CuO e. A metal that burns in oxygen with a bright white light. Ans. Magnesium is a metal that burns in oxygen with a bright white light. 2Mg + O2 → 2MgO Q#29 How will you show that copper is more reactive than silver? Ans. Copper displaces silver from silver nitrate solution. Therefore, copper is more reactive than silver

because more reactive metal displaces less reactive metal from its salt solution. Reaction: Cu + 2AgNO3 → Cu (NO3)2 + 2Ag Q#30 A set of metals in order of their increasing chemical reactivity is given below: silver, copper, lead, iron, zinc, magnesium and sodium. a. Which of the above metals is stored in kerosene? Ans. Sodium is stored in kerosene. b. Which metals will react with cold water? Ans. Sodium and magnesium will react with cold water. c. Which gas will be liberated when metals react with cold water? Ans. Hydrogen gas will be liberated when metals react with cold water. d. Which of the metals will react with oxygen when heated? Ans. Magnesium will react with oxygen when heated. e. Which of the metals becomes black in the presence of hydrogen sulphide H2S? Ans. Silver becomes black in the presence of hydrogen sulphide H2S. f. Which of the metals burns with white bright flame in oxygen? Ans. Magnesium burns with white bright flame in oxygen. Q#31 Illustrate with reference to four physical properties and one chemical property, the chief differences

between metals and non metals. For each physical property that you choose, note down at least one exception to the general statement.


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METALS NON-METALS

1. Metals usually have a silver or grey colour

except gold and copper.

2. All metals are hard except sodium and potassium which can be cut with a knife.

3. Metals usually exist in the solid form at

room temperature except mercury which exists as liquid at room temperature.

4. Metals usually have high density except

sodium and potassium which are even lighter than water.

5. Metals react with oxygen to form basic

oxides which dissolve in water to form bases. 2Mg + O2 → 2MgO MgO + H2O → Mg(OH)2

1. They occur in many colours, e.g. sulphur is

yellow and chlorine is yellowish green.

2. Non metals are not generally hard. Solid non-metals are brittle.

3. Non-metals may occur as solids, liquids or

gases at room temperature e.g. sulphur, carbon and iodine are solids. Bromine is a liquid while chlorine and nitrogen are gases.

4. Non-metals have low densities except iodine

which has high density.

5. Non metals react with water to give acidic oxides which dissolve in water to form acids.

S + O2 → SO2 SO2 + H2O → H2SO3

Q Why hydrogen gas been placed in the reactivity series? Ans. Though hydrogen is not a metal but still it has been placed in the reactivity series of metals

because hydrogen also loses electrons and forms positive ions H+ like metals and it acts as borderline between highly reactive and least reactive metals.

Q#2 Why some metals are more reactive and others less reactive? Ans. Reactivity of metals depends on the ease with which they lose electrons to form positive ions.

Now, if a metals losses electrons easily to form positive ions, it will react rapidly with other substance and hence it will be highly reactive metal. On the other hand, if a metal loses electrons less readily to form positive ions, it will react slowly with other substance. Such a metal will be less reactive.

Q Corrosion of Aluminium. Ans. Aluminium vessels lose their shine and become dull very soon after use. This is due to the

corrosion. The oxygen of air reacts with aluminium to form a thin, dull layer of Aluminium oxide (Al2O3) all over the vessel. This Aluminium oxide layer is very tough and prevents the metal underneath from further corrosion. This Aluminium oxide layer can be made thicker by electrolysis (to give them more protection from corrosion) and this process is known as Anodising.

Q Corrosion of Copper Ans. The copper objects lose their shine after sometime due to the formation of copper oxide layer on

them. When a copper object remains in damp air for a considerable time, then copper metal reacts slowly with CO2 and H2O of air to form a green coating of basic copper carbonate on the surface of an object. This basic copper carbonate is a mixture of copper carbonate and copper hydroxide CuCO3.Cu(OH)2.


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The corroded copper metals can be cleaned with dilute acid solution. Q Corrosion of silver Ans. When shining silver objects lose their shine and becomes dull, we say that it has been tarnished.

When silver objects are kept in air, they get tarnished and gradually turn black. Silver is a noble metal, so it does not react with oxygen present in air. But air has hydrogen sulphide gas (H2S). So, silver objects react with hydrogen sulphide gas to form a black coating of silver sulphide (Ag2S). Thus, silver objects turn black due to the formation of silver sulphide layer on their surface.

Uses Of Metals

Copper and Aluminium metals are used to make wires to carry electric current. This is because they have very low electrical resistance and hence are very good conductors of electricity.

Iron, copper and aluminium metals are used to make house-hold utensils and factory equipments.

Iron is also used for making bells etc. because of its sonorous property. Zinc is used for galvanizing iron to protect it from rusting. Chromium and nickel metals are used for electroplating and in the manufacture of stainless

steel. The Aluminium foils are used in packaging of medicines, cigarettes and food materials. Silver and gold metals are used to make jewellery. The liquid metal mercury is used in making thermometers.

Metal Reactivity Series:- From the action of oxygen, water and acid with metals, it is clear that some metals are more reactive than others. On the basis of rate of reaction of metals with oxygen, water and acids, the metals have been arranged in serial order or group according to their chemical reactivity.

The arrangement of metals in a vertical column in the decreasing order of their chemical reactivities is called metal activity series: Potassium (K) Sodium (Na)

Calcium (Ca)

Magnesium (Mg)

Aluminium (Al)

Zinc (Zn)

Iron (Fe)

Tin (Sn)

Lead (Pb)

Hydrogen (H)

Copper (Cu)

Mercury (Hg)

Silver (Ag)

Gold (Au)

Platinum (Pt)

Dec

reas

ing

Met

al R

eact

ivity


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Hydrogen is a non metal but it has been placed in metal reactivity series because hydrogen like metals can donate an electron to form H+ ions. It also acts as border line between more reactive and less reactive metals.

Metals more reactive than hydrogen:- The metals which can lose their valence electrons more readily than hydrogen are called more reactive metals. Eg. K, Na, Ca, Mg, Al, Zn, Fe, Sn and Pb.

Metals less reactive than Hydrogen:- The metals which lose their valence electrons with more difficulty than hydrogen are called less reactive metals e.g. Cu, Hg, Ag, Au and Pt.

Displacement Reaction It is a type of reaction in which highly reactive metal displaces less reactive metal from its salt solution. Examples of displacement reaction:-

a)Reaction of copper metal with silver nitrate solution:- When a strip of copper (red in colour) is placed in clear silver nitrate solution in a beaker, following observations are made. i) The clear silver nitrate solution gradually changes to blue solution. ii) The strip of copper gets coated with shining layer of silver. Reason:- When copper is placed in the solution of silver nitrate, copper being more reactive than silver, displaces silver from silver nitrate solution. Copper metal dissolves to form copper nitrate which is blue in colour. 2AgNO3 + Cu(s) Cu(NO3)2 + 2Ag Silver nitrate Copper nitrate (Colourless solution) (Blue solution)

b)Reaction of Iron metal with copper sulphate solution:-

When as strip of Iron metal is placed in copper sulphate solution, following observations are made:- i) The blue colour of copper sulphate gradually fades. ii) The colour of solution gradually changes to light green. iii) Reddish brown precipitate of copper is deposited on iron. Reason:- Iron metal is more reactive than copper when a strip of iron is placed in copper sulphate solution, iron being more reactive this places copper from copper sulphate solution. The iron metal dissolves to form ferrous sulphate. CuSO4(aq) + Fe(s) FeSO4(aq) + Cu(s) Exactly in the same way zinc metal displaces copper from copper sulphate solution.


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SUBJECT – SOCIAL STUDIES HISTORY- CHAPTER 6: Weavers, Iron Smelters and Factory Owners

A. Short answer types questions: Q1. What kind of cloth had a large market in Europe? Ans. The printed cotton textiles had a large market in Europe mainly for their exquisite floral design, fine texture and for relative cheapness. Q2. What is Jamdani? Ans. Jamdani I a fine muslin on which decorative motifs are woven on the loom, typically in grey and white. Often a mixture of cotton and gold thread was used in it. The most important centres of Jamdani weaving were Decca in Bengal and Lucknow in the United provinces. Q3. What is Bandana? Ans. The word bandana is originally derived from the word “bandhna” and referred to a variety of brightly coloured cloth produced. Though a method of tie and dye. The word bandana now refers to any brightly coloured and printed scarf for the neck or head. Q4. Who are the Agaria? Ands. A community which earned their living by smelting of iron are known as the agaria. Q5. Fill in the blanks:

1. The word chintz comes from the word chhint. 2. Tipu’s sword was made of wootz steel. 3. India’s textile exports declined in the end of the eighteenth century. B. Long answer type questions:

Q1. The names of different textiles tell us about their histories. How? Ans. From the sixteenth century, European trading companies began buying Indian textiles for sale in Europe. The European traders first encountered fine cotton cloth from India carried by Arab merchants in Mosul in present day Iraq. So, they began referring to all finely woven textiles as “muslin” a word that acquired wide currency. Similarly, the Portuguese, landed in calicut took back the cotton textiles which came to be called “calico” (derived from calicut) and subsequently calico became the general name for all cotton textiles.

Q2. Why did the wool and silk producers in England protest against the import of Indian textiles in the early eighteenth century? Ans. By the early eighteenth century, the wool and silk producers in England began protesting against the import of Indian textiles due to the popularity of Indian textiles in Europe especially in England. The English textiles were unable to compete with Indian textiles. So, the English producers wanted a secure market within the country by preventing the entry of Indian textiles. The first to grown under government protection was the calico printing industry. Indian designs were now initiated and printed in England white muslin Indian cloth.

Q3. How did the development of cotton industries in Britain affect textile producers in India? Ans. The development of cotton industries in Britain affect the textiles in India in several ways:

1. Indian textiles now had to compete with Britain in European and American markets. 2. Vey high duties were imposed on Indian textiles making its import into Britain difficult.


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3. With the beginning of the nineteenth century, English made cotton textiles successfully ousted Indian textiles from their traditional markets in Africa, America and Europe.

4. By the 1830s, British cotton cloth flooded the Indian markets and two thirds of all the cotton clothes worn by Indians were made of cloth produced in British.

5. Thousands of weavers in India were now thrown out of employment because, the English and European companies stopped buying Indian goods and their agents no longer gave out advances to weavers to secure supplies.

Q4. Why did the Indian iron smelting industry decline in the nineteenth century? Ans. The Indian iron smelting industry declined due to various reasons in the nineteenth century. They are as:

1. In most villages, furnaces fell into disuse and the amount of iron produced cane down by the late nineteenth century

2. The new forest las of colonial government prevented people from entering the reserved forests, made difficult to the iron smelters find wood for charcoal.

3. If the government did grant access to forests the iron smelters had to pay a very high tax to the forest department for very furnace. They used this to reduce their income.

4. Moreover, by late nineteenth century iron and steel was being imported from Britain. Iron-smelters in India began using the imported iron to manufacture utensils and implements. This inevitably lowered the demands for iron produced by local smelters.

Q5. What problems did the Indian textile industry face in the early years of its development? Ans. In the first few decades of its existence, the Indian textile industry faced many problems. It found it difficult to compete with the cheap textiles imported from Britain. The colonial government in India usually refused to impose heavy duties on imports from other countries. There was lack of credit facilities under the British rule and lack of protection to Indian industry against unfair competition of foreign goods. The first major spurt in the development of cotton of cotton factory production in India, therefore was during the first world war when textile imports from Britain declined and Indian factories were called upon to produce cloth for military supplies.

Q6. Find out the cause of expansion of TISCO steel production during the first world war? Ans. The Tata Iron and Steel company (TISCO) began producing steel in 1917. By the first world war steel produced in Britain had to meet demands of war in Europe. So, imports of British steel into India declined dramatically and the Indian railways turned to TISCO for supply of rails. As the war dragged on for several years, TISCO had to produce shells and carriage wheels for the war. By 1919, the colonial government was buying 90% of the steel manufactured steel industry within the British Empire.

CHAPTER 7: Civilising the ‘Native’, Educating the Nation A) Answer the following:

Q1. Why did William Jones feel the need to study Indian history, philosophy and law? Ans. William Jones started studying Indian ancient text on law, history and philosophy to discover the ancient Indian heritage, mastering Indian languages and translating Sanskrit and Persian works in English. He gave a deep respect for the ancient cultures of India and the west by discovering ancient texts, understanding their meaning, translating them and make their finding known to others, Jones believed that it would not only help the British to learn from Indian culture, but it would also help Indians rediscover


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their own heritage and understand the lost glories of their pasts. In this process the British would became the guardians of Indian culture as well as its masters.

Q2. Why did James Mill and Thomas Macaulay think that European education was essential in India? Ans. James Mill and Thomas Macaulay began to criticise the orientalist vision of learning. To them knowledge of the east was full of errors and unscientific thought. They argued Indians should be made familiar with the scientific and technical advances that the west has made, so as to teach what was useful and practical. They saw India as an uncivilised country that needed to be civilised and no branch of eastern knowledge could be compounded to what England had produced. They felt that European education make Indians aware of the developments in western science and philosophy. The European learnings would enable Indians to recognise the advantages of the expansion of trade and commerce, and make them see the importance of developing the resources of a country. The European education could be way of civilising the people, changing their tastes, values and culture

Q3. Why did Mahatma Gandhi want to teach children handicrafts? Ans. Mahatma Gandhi said that education ought to develop a person’s mind and soul. Literacy or simply learning to read and write, by itself did not count to education. He argued that people had to work with their hands, learn a craft and know how different things operate. This would develop their mind and capacity to understand. So, he wanted to teach children handicrafts rather than reading and writing textbooks. He also argued child’s education should begin by teaching it a useful handicraft and enabling it to produce from the moment it begins its training.

Q4. Why did Mahatma Gandhi think that English education had enslaved Indians? Ans. Mahatma Gandhi argued that the colonial education created a sense of inferiority in the minds of Indians. It made them see western civilisation as superior and destroyed the pride they had in their own culture. There was poison in their education, it enslaved Indians and it cast an evil spell on them. The Indians educated in these institutions were charmed by the west and being admiring British rule. He also believed that education in English crippled Indians, distanced them from their own social surroundings and made them “strangers in their own land”. It also dispersed their local cultures.

B) State whether true or false: 1. Asiatic society was founded by James Mill. 2. The 1854 despatch on education was in favour of English being introduced as a medium of

higher education in India. 3. Mahatma Gandhi favoured western education. 4. Rabindranath Tagore felt that children ought to be subjected to strict discipline.

Ans. a) False, b) False, c) False, d) True.

CHAPTER 8: Women, Caste and Reform A) Answer the following questions:

Q1. Which social ideas did the following people support: Ans.1. Rammohun Roy: Raja Rammohun Roy founded a reform association known as the Brahmo Sabha (later known as the Brahmo Samaj) in Calcutta. He was keen to spread the knowledge of western education


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in the country and bring about greater freedom and equality for women. He was against the practise of sati. 2. Dayanand Saraswati: Swami Dayanand Saraswati founded the reform association called Arya Samaj. He was in favour of widow marriage. He advocated that there was no place for social evils like caste system, child marriages and untouchability in the society. He was against the caste system on the Hindu society. 3. Veersalingam Pantulu: Kandukuri Veersalingam Pantulu was an outstanding leader of the reform movements in southern India. He dedicated himself to the cause of social reform. He worked for the enlightenment of the people. His greatest contribution was favouring the cause of emancipation of women. This included promoting girl education and widow marriages. 4. Jyotirao Phule: Jyotirao Phule founded the Satyashodhak Samaj. He proposed that shudras and anti-shudras should unite to challenge caste discrimination. He was concerned about the plight of upper caste women, the miseries of the labourer and the humiliation of the lower caste. 5. Pandita Ramabai: Pandita Ramabai ws a great scholar of Sanskrit. She founded a widow’s home at Pune to provide shelters to widows who had been treated badly by their husband’s relative

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