Category: Class 8

Students can Download Maths Chapter 1 Numbers Ex 1.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.3

Question 1.
Fill in the blanks:
(i) The ones digits in the cube of 73 is……….
(ii) The maximum number of digits in the cube of a two digit number is……….
(iii) The cube root of 540 × 50 is………..
(iv) The cube root of 0.000004913 is………..
(v) The number to be added to 3333 to make it a perfect cube is………….
Solution:
(i) 7
(ii) 6
(iii) 30
(iv) 0.017
(v) 42

Question 2.
Say True or False:
(i) The cube of 0.0012 is 0.000001728.
(ii) The cube root of 250047 is 63.
(iii) 79570 is not a perfect cube.
Solution:
(i) false
(ii) true
(iii) true

Question 3.
Show that 1944 is not a perfect cube.
Solution:

1994 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 2³ × 3³ × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.

Question 4.
Find the cube root of 24 × 36 × 80 × 25.
Solution:

Question 5.
Find the smallest number by which 10985 should be divided so that the quotient is a perfect cube.
Solution:

We have 10985 = 5 × 13 × 13 × 13
= 5 × 13 × 13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.

Question 6.
Find two smallest perfect square numbers which when multiplied together gives a perfect cube number.
Solution:
Consider the numbers 22 and 42
The numbers are 4 and 16.
Their product 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16

Question 7.
If the cube of a squared number is 729, find the square root of that number.
Solution:
729 = 3 × 3 × 3 × 3 × 3 × 3
(729)^1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]

we have to find out √3, √3 = 1.732

Question 8.
What is the square root of cube root of 46656?
Solution:
We have to find out \(\sqrt{(\sqrt[3]{46656})}\)
First we will find \(\sqrt[3]{46656}\)

\(\sqrt[3]{46656}\) = (2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3)^1/3
\(\sqrt[3]{46656}\) = 2 × 2 × 3 × 3
\(\sqrt[3]{46656}\) = 2² × 3² = 36
Now \(\sqrt{(\sqrt[3]{46656})}\) = \(\sqrt{36}\) = \(\sqrt{2^2×3^2 }\) = 2 × 3 = 6
∴ The required is 6.

Question 9.
Find the cube root of 729 and 6859 by prime factorisation.
Solution:

(i) \(\sqrt[3]{729}\) = \(\sqrt[3]{3 × 3 × 3 × 3 × 3 × 3}\)
= 3 × 3
\(\sqrt[3]{729}\) = 9

(ii) \(\sqrt[3]{6859}\) = \(\sqrt[3]{19 × 19 × 19 }\)
\(\sqrt[3]{6859}\) = 19

Question 10.
Find the smallest number by which 200 should be multiplied to make it a perfect cube.
Solution:
We find 200 = 2 × 2 × 2 × 5 × 5
Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.

200 × 5 = (2 × 2 × 2 × 5 × 5) × 5
1000 = 2 × 2 × 2 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.

Students can Download Maths Chapter 2 Life Mathematics Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Ex 2.1

Question 1.
Fill in the blanks:
(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job together is………..days.
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in………..days.
(iii) A can do a work in 24 days. A and B together can finish the work in 6 days. Then B alone can finish the work in…………days.
(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in………..days.
(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for Rs 200000. The amount that A will get is………
Solution:
(i) 2 days
(ii) 5
(iii) 8
(iv) 25
(v) Rs 1,20,000

Question 2.
210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Solution:
Let the required number of men be x.

More working hours ⇒ less men required.
∴ It is inverse proportion.
∴ Multiplying factor is \(\frac{12}{14}\)
Also more number of days ⇒ less men
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{18}{20}\)
∴ x = 210 × \(\frac{12}{14}\) × \(\frac{18}{20}\)= 162 men

162 men are required.

Question 3.
A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?
Solution:
Let the required number of cement bags be x.

Number of days more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{18}{12}\)
Number of machines more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{24}{36}\)
∴ x = 7000 × \(\frac{18}{12}\) × \(\frac{24}{36}\)

x = 7000 cement bags
7000 cement bags can be made.

Question 4.
A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 hours more a day?
Solution:
Let the required number of days be x.

To produce more soaps more days required.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{14400}{9600}\)
If more hours spend, less days required.
∴ It is indirect proportion.
∴ Multiplying factor = \(\frac{15}{18}\)
∴ x = 6 × \(\frac{14400}{9600}\) × \(\frac{15}{18}\)

x = \(\frac{15}{2}\)
\(\frac{15}{2}\) days will be needed.

Question 5.
If 6 container lorries transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?
Solution:
Let the number of lorries required more = x.

As the goods are more ⇒ More lorries are needed to transport.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{180}{135}\)
Again if more days ⇒ less number of lorries enough.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{5}{4}\)
∴ 6 + x = 6 × \(\frac{180}{135}\) × \(\frac{5}{4}\)

6 + x = 10
x = 10 – 6
x = 4
∴ 4 more lorries are required.

Question 6.
A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?
Solution:
Time taken by A to complete the work =12 hrs.
∴ A’s 1 hr work = \(\frac{1}{12}\)…………(1)
(B + C) complete the work in 3 hrs.
∴ (B + C)’s 1 hour work = \(\frac{1}{3}\)…………(2)
(1) + (2) ⇒
∴ (A + B + C)’s 1 hour work = \(\frac{1}{12}\) + \(\frac{1}{3}\) = \(\frac{1+4}{12}\) = \(\frac{5}{12}\)
Now (A + C) complete the work in 6 hrs.
∴(A + C)’s 1 hour work = \(\frac{1}{6}\)
∴ B’s 1 hour work = (A + B + C)’s 1 hour work – (A + C)’s 1 hour work
= \(\frac{5}{12}\) – \(\frac{1}{6}\) = \(\frac{5-2}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
∴ B alone take 4 days to complete the work.

Question 7.
A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?
Solution:
(A + B) complete the work in 12 days.
∴ (A + B)’s 1 day work = \(\frac{1}{12}\)……….(1)
(B + C) complete the work in 15 days
∴ (B + C)’s 1 day work = \(\frac{1}{15}\)……….(2)
(A + C) complete the work in 20 days
∴ (A + C)’s 1 day work = \(\frac{1}{20}\)……….(3)
Now (1) + (2) + (3) ⇒
[(A + B) + (B + C) + (A + C)]’s 1 day work = \(\frac{1}{12}\) + \(\frac{1}{15}\) + \(\frac{1}{20}\)

(2A + 2B + 2C)’s 1 day work = \(\frac{5}{60}\) + \(\frac{4}{60}\) + \(\frac{3}{60}\)
2(A + B + C)’s 1 day work = \(\frac{5+4+3}{60}\)
(A + B + C)’s 1 day work = \(\frac{12}{60×2}\)
(A + B + C)’s 1 day work = \(\frac{1}{10}\)
Now A’s 1 day’s work = (A + B + C)’s 1 day work – (B + C)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{15}\) = \(\frac{3}{30}\) – \(\frac{2}{30}\) = \(\frac{1}{30}\)
∴ A takes 30 days to complete the work.
B’s 1 day work = (A + B + C)’s 1 day’s work – (A + C)’s 1 day’s work
= \(\frac{1}{10}\) – \(\frac{1}{20}\) = \(\frac{6}{60}\) – \(\frac{3}{60}\)
\(\frac{6-3}{60}\) = \(\frac{3}{60}\) = \(\frac{1}{20}\)
B takes 20 days to complete the work.
C’s 1 day work = (A + B + C)’s 1 day work – (A + B)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{12}\) = \(\frac{6}{60}\) – \(\frac{5}{60}\) = \(\frac{6-5}{60}\) = \(\frac{1}{60}\)
∴ C takes 60 days to complete the work.

Question 8.
Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 more minutes than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Solution:
Time taken by A to fit a chair = 15 minutes
Time taken by B = 3 minutes more than A
= 15 + 3 = 18 minutes
∴ A’s 1 minute work = \(\frac{1}{15}\)
B’s 1 minute work = \(\frac{1}{18}\)
(A + B)’s 1 minutes work = \(\frac{1}{15}\) + \(\frac{1}{18}\)

\(\frac{12}{180}\) + \(\frac{22}{180}\) = \(\frac{22}{180}\) = \(\frac{11}{90}\)
∴ Time taken by (A + B) to fit a chair
= \(\frac{1}{\frac{11}{90}}\) = \(\frac{90}{11}\) minutes
∴ Time taken by (A + B) to fit 22 chairs
= \(\frac{90}{11}\) × 22 = 180 minutes
= \(\frac{180}{60}\) = 3 hours

Question 9.
A man takes 10 days to finish a job where as a woman takes 6 days to finish the same job. Together they worked for 3 days and then the woman left. In how many days will the man complete the remaining job?
Solution:
Man can finish the work in 10 days and women can finish the same work in 6 days.
∴ Man’s 1 day work = \(\frac{1}{10}\)
Woman’s 1 day work = \(\frac{1}{6}\)
(Man + Woman)s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{6}\) = \(\frac{6}{60}\) + \(\frac{10}{60}\) = \(\frac{16}{60}\)
(Man + Woman)s 3 days work

In 3 days \(\frac{4}{5}\) th of the whole work is completed.
Remaining work = 1 – \(\frac{4}{5}\) = \(\frac{5}{5}\) – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Complete work is done by the man by 10 days
∴ \(\frac{1}{5}\) of the work is done by man in \(\frac{1}{5}\) × 10 = 2 days.

Question 10.
A is thrice as fast as B. If B can do a piece of work in 24 days then, find the number of days they will take to complete the work together.
Solution:
If B does the work in 3 days, A will do it in 1 day.
B complete the work in 24 days.
∴ A complete the same work in \(\frac{24}{3}\) = 8 days.
∴ (A + B) complete the work in \(\frac{ab}{a+b}\) days = \(\frac{24×8}{24+8}\) days = \(\frac{24×8}{32}\)days = 6 days

They together complete the work in 6 days.

Students can Download Maths Chapter 1 Life Mathematics Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Intext Questions

Exercise 1.1
Try This (Text book Page no. 1)

Question 1.
Find the indicated percentage value of the given numbers
Solution:

Think (Text book page no. 6)

Question 1.
An increase from 200 to 600 ¡s clearly a 200%increase. Isn’t it? (check!). With a lot of pride, the traffic police commiuêener of a city reported that the accidents had decreased by 200% ¡n one year. He cameupwith this number stating that the accidents had gone down from 600 last year to 2this year. Is the decrease from 600 to 200, the same 200% as above? Justify.
Solution:
Increase from original value 200 to 600

Decrease from original value 600 to 200

here original value is 600
% decrease = \(\frac{600-200}{600}\) x 100 = \(\frac{400}{600}\) x 100 = 66.67% decrease
Increase from 200 → 600 and % decrease from 600 → 200 are not the same

Try This (Text book page no. 7)

Question 1.
What percent of a day is 10 hours?
Solution:
In a day, there are 24 hours
10 hrs out of 24 hrs is \(\frac{10}{24}\)
As a percentage, we need to multiply by 100
∴ Percentage = \(\frac{10}{24}\) x 100 = 41.67%

Question 2.
Divide ₹ 350 among P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.
Solution:
Let R get x, Q gets 50% of what R gets
∴ Q gets = \(\frac{50}{100}\) × x = \(\frac{x}{2}\)
P gets 50% of what Q gets
∴ P gets = \(\frac{50}{100}\) x \(\frac{x}{2}\) = \(\frac{x}{4}\)
Since 350 is divided among the three
∴ 350 = x + \(\frac{x}{2}\) + \(\frac{x}{4}\)
350 = \(\frac{4x+2x+x}{4}\) = \(\frac{7x}{4}\) = 350
x = \(\frac{350×4}{7}\) = 200
Q gets = \(\frac{x}{2}\) = \(\frac{200}{2}\) = 100,
P gets = \(\frac{x}{4}\) = \(\frac{200}{4}\) = 50
∴ P = 50
Q = 100
R = 200

Exercise 1.2
Think (Text book Page No. 13)

Question 1.
A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?
Solution:
Let cost price of marker board be 100
CP = 100 Marks it 15% above CP
∴ Marked price = MP = \(\frac{15}{100}\) x CP + CP = \(\frac{15}{100}\) 100 + 100 = 15 + 100 = 115
Discount % = 15%

∴ He sells it 97.75 which is less than his cost price. Therefore he loses
Loss = 97.75 – 100 = – 2.25

Try This (Text book Page No. 14)

Question 1.
By selling 5 articles, a man gains the cost price of 1 article. Find his gain percentage.
Solution:
Let cost price of article be C.P. Let S.P of 1 articles be SP by selling 5 articles at SP he makes a gain of cost price of one article.
Gain on 1 article = SP – CP; ⇒ Gain% = \(\frac{SP-CP}{CP}\) x 100
Gain on 2 articles = 2 x (SP – CP)
Gain on 5 articles = 5 x (SP – CP)
Given than gain on 5 articles is CP of 1 article
∴ 5(SP – CP) = CP
\(\frac{SP-CP}{CP}\) = \(\frac{1}{5}\)
Gain percentage \(\frac{SP-CP}{CP}\) x 100 = \(\frac{1}{5}\) x 100 % = 20%

Question 2.
By selling 8 articles, a shopkeeper gains the selling price of 3 articles. Find his gain percentage.
Solution:
Let cost price of 1 article be CP. Let selling price of 1 article be SP.
Gain on 1 article = SP – CP
Gain on 8 articles = 8 x (SP – CP)
Given that gain on 8 articles in selling price of 3 articles
8(SP – CP) = 3 x SP
∴ \(\frac{SP-CP}{CP}\) = \(\frac{3}{8}\)
∴ \(\frac{SP}{SP-CP}\) = \(\frac{8}{3}\)
Subtracting 1 on both sides
\(\frac{SP}{SP-CP}\) – 1 = \(\frac{8}{3}\) – 1 = \(\frac{SP-(SP-CP)}{SP-CP}\)= \(\frac{8-3}{3}\)
\(\frac{SP-SP-CP}{SP-CP}\) = \(\frac{5}{3}\) ⇒ \(\frac{CP}{SP-CP}\) = \(\frac{5}{3}\)
\(\frac{SP-CP}{CP}\) = \(\frac{3}{5}\) (taking reciprocals on both sides)
Gain% \(\frac{SP-CP}{CP}\) x 100 = \(\frac{3}{5}\) x 100 = 3 x 20 = 60%

Question 3.
If the C.P of 20 articles is equal to the S.P of 15 articles, find the profit or loss percentage.
Solution:
Given CP of 20 article = SP of 15 articles
∴ SP of 15 articles = CP of 20 articles
∴ SP of 1 article = \(\frac{1}{15}\) x CP of 20 articles
SP = \(\frac{1}{15}\) x 20 CP = \(\frac{20}{15}\) CP = \(\frac{4}{3}\) CP
∴ SP = \(\frac{4}{3}\) CP ⇒ SP is greater than CP
It is a profit.

Exercise 1.3
Try This (Text book Page No. 23)

Question 1.
Find the principal which gives ₹ 420 as C.I @ 20% p.a compounded half yearly for one year.
Solution:
CI = ₹420
Rate = ₹ 20% p.a
Principle = ₹ [required to find] Time period (n) = 1 year.
However, let us value of r to be 20% p.a so for half yearly, r is \(\frac{20}{2}\) = 10%
Formula for Amount (A) when compounded half yearly is

Question 2.
The price of a laptop depreciates @ 4% p.a. If its present price is ₹ 24,000, find its price after 3 years.
Solution:
Let original price of laptop be ‘P’, Rate of depreciation is 4% p.a,
Present price is ₹ 24,000 (D).
Formula for depreciation is

Price after 3 years from now is
P(1 – \(\frac{4}{100}\))ⁿ⁺³ = ? ⇒ (1 – \(\frac{4}{100}\))ⁿ (1 – \(\frac{4}{100}\))³
From (1)
24,000 x (1 – \(\frac{4}{100}\))³
24,000 x \(\frac{96}{100}\) x \(\frac{96}{100}\) x \(\frac{96}{100}\) = 21233.66

Actvity I (text book Page No. 23)

Question 1.
Mukunthan invests 30,000/- for 3 months in a bank which gives C.I at the rate of 12% compounded monthly. A private company offers his S.l at the rate of 12% p.a What is the difference in the interests received by Mukunthan? Do by traditional method and verify your answer by calculator.
Solution:
Principal = 30,000
Time period = 3 months
In Bank rate of interest for CI = 12% compounded monthly
∴ A = (1 + \(\frac{r}{100}\))ⁿ = 30,000(1 + \(\frac{12}{100}\))³
30,000 x \(\frac{112}{100}\) x \(\frac{112}{100}\) x \(\frac{112}{100}\) = 42147.84
∴ CI = A – P = 42147.84 – 30,000
CI = 12147.84
In private company,
Rate of single Interest SI = 12% p.a
So, for 3 months, i.e \(\frac{3}{12}\) = \(\frac{1}{4}\) year,

∴ Difference in interest = CI – SI = 12,147.84 – 900 = 11247.84

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Students can Download Maths Chapter 1 Numbers Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Additional Questions

Question 1.
Find the least number by which 1100 must be multiplied so that the product becomes a perfect square. Also, in each case find the square root of the perfect square so obtained.
Solution:
We find 1100 = 2 × 2 × 5 × 5 × 11 =2² × 5² × 11
∴ The prime factor 11 has no pair.
∴ If we multiply 1100 by 11, then the product becomes a perfect square.
∴ New number = 1100 × 11 = 12100

12100 = 2² × 5² × 11²
\(\sqrt{12100}\) = 2 × 5 × 11 = 110
Square root of the new number = 110

Question 2.
Find the square of 509 using (a + b)² = a² + 2ab + b²
Solution:
509² = (500 + 9)² = 500² + 2 x 500 x 9 + 9²
= 250000 + 9000 + 81
509² = 259081

Question 3.
Find the sum without adding
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Given sum is the sum of first 12 odd natural numbers.
Sum of first n odd natural numbers is n².
∴ 1 +3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12² = 144

Question 4.
Write a Pythagorean triplet whose one number is 110
Solution:
Here let 2m = 110
m = 55

m² – 1 = 55² – 1 = 3025 – 1 = 3024
m² + 1 = 55² + 1 = 3025 + 1 = 3026
∴ Pythagorean triplet is 110, 3024, 3026.

Question 5.
Find the square root of 10 \(\frac{2}{3}\) correct to three places of decimal.
Solution:
10 \(\frac{2}{3}\) = 10.6666…….

\(\sqrt{10 \frac{2}{3}}=3.2659 \Rightarrow \sqrt{10 \frac{2}{3}}\) = 3.266 correct to three places of decimal.

Question 6.
Find the square root of 0.053361
Solution:

\(\sqrt{0.053361}\) = 0.231

Question 7.
Three numbers are in the ratio 2:3:4. The sum of their cubes is 33957. Find the numbers.
Solution:
Let the numbers be 2x, 3x and 4x
(2x)³ + (3x)³ + (4x)³ = 33957
8x³ + 27x³ + 64x³ = 33957
99x³ = 33957
x³ = \(\frac{33957}{99}\)
x³ = 343
x³ = 7 × 7 × 7
x³ = 7³
x = 7
∴ The numbers are 2x = 2 × 7 = 14
3x = 3 × 7 = 21
4x = 4 × 7 = 28

Question 8.
The volume of a cube is 9261000 m³. Find the side of the cube?
Solution:
Volume of the cube = side x side x side
side x side x side = 9261000
side = \(\sqrt[3]{9261×1000}\) = \(\sqrt[3]{9261}\) × \(\sqrt[3]{1000}\)
= \(\sqrt[3]{3^{3}×7^{3}}\) × \(\sqrt[3]{10×10×10}\) = 3 × 7 × 10 = 210
∴ Side of the cube = 210 m

Question 9.
If the diameters of the sun and Earth are 1.4 × 109 m and 1.275 × 107 m respectively. compare these two.
Solution:

So the diameter of the Sun is about 100 times the diameter of the Earth.

Question 10.
The size of a red blood cell is 0.000007 m and that of a plant, cell is 0.00001275 m. Compare these two.
Solution:

∴ RBC size if half the size of a plant cell.

Students can Download Science Term 3 Chapter 9 Visual Communication Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Science Solutions Term 3 Chapter 9 Visual Communication

Samacheer Kalvi 8th Science Visual Communication Text Book Exercise

I. Choose the best answer:

Question 1.
The Keyboard shortcut is used to copy the selected text
(a) Ctrl + C
(b) Ctrl + V
(c) Ctrl + X
(d) Ctrl + A
Answer:
(a) Ctrl + C

Question 2.
The Keyboard shortcut is used to cut the selected text
(a) Ctrl + C
(b) Ctrl + V
(c) Ctrl + X
(d) Ctrl + A
Answer:
(c) Ctrl + X

Question 3.
If the ruler is not displayed in the screen, option is clicked.
(a) View → ruler
(b) view → task
(c) File → save
(d) Edit → paste
Answer:
(a) View → ruler

Question 4.
How many types of page orientation are there in Libre office Writer?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 5.
The menu used to save the document is –
(a) File → open
(b) file → print
(c) file → save
(d) Edit → close
Answer:
(c) file → save

II. Answer briefly:

Question 1.
What is the use for Text document software?
Answer:

1. A text file is used to store standard and structured textual data or information that is human readable.
2. It is defined in several different formate including the most popular ASCII for cross platform usage and ANSI for windows – based operating platforms.

Question 2.
What is selecting text?
Answer:
Selecting is the process of highlighting text or picking an object. For example, a user may select text to copy, cut or move that text to an alternate location or select a file they want to view.

Question 3.
How to close a document?
Answer:
Close the current document by selecting File → Close command on the menu bar or click the Close icon if it is visible on the Standard tool bar.

Question 4.
What is right alignment?
Answer:
Right alignment is text or page formatting that aligns text along the right side of a page or containing element.

Question 5.
How to open an existing document?
Answer:
To open an existing document, do any one of the following methods:

1. Click the Open File button on the menu bar.
2. Choose File → Open command from the menu bar.
3. Press CTRL+O keys on the keyboard. Each of the above method will show the Open dialog box. Choose the file and click the Open button.

Samacheer Kalvi 8th Science Visual Communication Additional Questions

I. Choose the correct answer:

Question 1.
………….. is a powerful and free office suite, used by millions of people.
(a) LibreOffice
(b) Microsoft window
(c) JAVA
(d) HTML
Answer:
(a) LibreOffice

Question 2.
………….. can create and edit forms, views and relations.
(a) Calc
(b) Impress
(c) Base
(d) Math
Answer:
(c) Base

Question 3.
………….. is the LibreOffice formula or equation editor.
(a) Impress
(b) Drawing
(c) Base
(d) Math
Answer:
(d) Math

Question 4.
The menu is used to print the document.
(a) File → open
(b) File → print
(c) File → save
(d) File → close
Answer:
(b) File → print

Question 5.
A ………….. is a set of characters and numbers in a certain style.
(a) Font
(b) Bullets
(c) Underline
(d) Paragraph
Answer:
(a) Font

Question 6.
………….. alignment refers to the appearance of the left and right sides of the Paragraph.
(a) Right
(b) Letf
(c) Paragraph
(d) None
Answer:
(c) Paragraph

Question 7.
How many types of alignment can be selected in LibreOffice?
(a) two
(b) three
(c) four
(d) five
Answer:
(c) four

Question 8.
A ………….. orientation means a horizontal display.
(a) Landscape
(b) Portrait
(c) Both (a) and (b)
(d) None of these
Answer:
(a) Landscape

Question 9.
A page is shorter in height but wider in width
(a) Landscape
(c) Both (a) and (b)
(b) Portrait
(d) None of these
Answer:
(b) Portrait

II. Answer the following question:

Question 1.
What is drawing?
Answer:
Draw is a vector drawing tool that can produce everything from simple diagrams or flowcharts to 3D artwork.

Question 2.
How can you create a new document?
Answer:
To create a new document, do any one of the following methods

1. Click the New Document button on the menu bar.
2. Choose File → New command from the menu bar.
3. Press CTRL+N keys on the keyboard.

Question 3.
How can you print a document?
Answer:
To print a document or selected pages follow the steps given below:

1. Open the document to be printed.
2. Choose File → Print command on the menu bar. The Print dialog box will open. Select the Options like print range, Number of copies, Printer name etc. See that printer is switched on and the paper is available in the printer tray.
3. Click OK.

Question 4.
What is the difference between cut and copy?
Answer:
The main difference between Cut and Copy is that cut removes the selected data from its original position while copy creates a duplicate of the original content.

Question 5.
What is font?
Answer:
A font is a set of characters and numbers in a certain style. Each font looks different from other fonts.

Question 6.
What is paragraph alignment?
Answer:
Paragraph alignment refers to the appearance of the left and right sides of the paragraph.

III. Paragraph Questions:

Question 1.
What are the components of LibreOffice?(any five)
Answer:
LibreOffice includes the following components.

Text Document:
Writer is a featurerich tool for creating letters, books, reports, newsletters, brochures, and other documents.

Calc (spreadsheet):
Calc has all of the advanced analysis, charting, and decision making features expected from a high-end spreadsheet. It includes over 300 functions for financial, statistical and mathematical operations, among others.

Impress (presentations):
Impress provides all the common multimedia presentation tools, such as special effects, animation, and drawing tools.

Base (database):
Base provides tools for day-to-day database work within a simple interface. It can create and edit forms, reports, queries, tables, views, and relations, so that managing a relational database is much the same as in other popular database applications.

Math (formula editor):
Math is the LibreOffice formula or equation editor. You can use it to create complex equations that include symbols or characters not available in standard font sets.

Question 2.
How can you selecting the text?
Answer:
For selecting text, the mouse or the keyboard can be used.
Selecting Text with Mouse – Following steps are to be followed:

1. Insertion point is moved to the start of the text to be selected.
2. The left mouse button should be clicked, held down and dragged across the text to be selected.
3. When the intended text is selected, the mouse button should be released.

Selecting Text with Keyboard – Following are the steps to be followed:

1. Insertion point is moved to the start of the text to be selected.
2. The Shift key is pressed down and the movement keys are used to highlight the required text.
3. When the Shift key is released, the text is selected.

Question 3.
List the steps of moving the text.
Answer:
The selected text can be easily cut and pasted in the required location. Following steps are to be followed.

1. The text to be moved to a new location is selected.
2. Edit → Cut is selected or in the tool bar is selected to cut the selected text.
3. Insertion point is moved to the place where the text is to be pasted.
4. Edit → Paste is selected or in the tool bar is selected to paste the text in the new location. The text can also be pasted in this way to another or another type of document.

The following keyboard shortcuts can be used to move text.

• Ctrl + X → to Cut,
• Ctrl + V → to Paste

Question 4.
How can you change the margins?
Answer:
If the user is not having the exact value for the margins then the Ruler option on the View menu can be used to change the margins.
Following steps are used in this method:

1. If the ruler is not displayed in the screen, View → Ruler option is clicked.
2. The gray area of the ruler indicates the margin’s top area.
3. The mouse pointer is then moved in between the gray and white area of the ruler.
4. When the pointer is in the right spot, it changes into a line with arrows on both sides.
5. The margin guide is dragged to a new location.

Students can Download Maths Chapter 2 Life Mathematics Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Intext Questions

Exercise 2.1
Try These (Text book Page No. 33)

Classify the given examples as direct or inverse proportion:
(i) Weight of pulses to their cost.
Solution:
As weight increases cost also increases.
∴ Weight and cost are direct proportion.

(ii) Distance travelled by bus to the price of ticket.
Solution:
As the distance increases price to travel also increases.
∴ Distance and price are direct proportion.

(iii) Speed of the athelete to cover a certain distance.
Solution:
As the speed increases, the time to cover the distance become less.
So speed and time are in indirect proportion.

(iv) Number of workers employed to complete a construction in a specified time.
Solution:
As the number of workers increases, the amount of work become less, so they are in indirect proportion.

(v) Volume of water flown through a pipe to its pressure.
Solution:
As the pressure increases, volume also increases.
∴ They are direct proportions.

(vi) Area of a circle to its radius.
Solution:
If the radius of the circle increases its area also increases.
∴ Area and radius of circles are direct proportion.

Use the concept of direct and inverse proportions and try to answer the following questions:
Question 1.
A student can type 21 pages in 15 minutes. At the same rate, how long will it take the student to type 84 pages?
Solution:
Direct proportion
No. of minutes = x
k = \(\frac{21}{15}\)
\(\frac{21}{15}\) = \(\frac{84}{x}\)

Question 2.
The weight of an iron pipe varies directly with its length. If 8 feet of an iron pipe weighs 3.2 kg, find the proportionality constant k and determine the weight of a 36 feet iron pipe.
Solution:

Weight of 36 feet iron pipe = x
\(\frac{36}{x}\) = 2.5

Question 3.
A car covers a distance of 765 km in 51 litres of petrol. How much distance would it cover in 30 litres of petrol?
Solution:
Direct proportion
k = \(\frac{51}{765}\)
Distance cover = x km
\(\frac{30}{x}\) = \(\frac{51}{765}\)

Question 4.
If x and y vary inversely and x = 24 when y = 8, find x when y = 12.
Solution:
k = xy = 24 × 8 = 192
∴ 12 × x = 192

Question 5.
If 35 women can do a piece of work in 16 days, in how many days will 28 women do the same work?
Solution:
Inverse proportion
No. of days = x
k = 35 × 16
∴ 28 × x = 35 × 16

Question 6.
A farmer has food for 14 cows which can last for 39 days. How long would the food last, if 7 more cows join his cattle?
Solution:
Inverse variation
k = xy = 14 × 39
No. of cow = 14 + 7 = 21
No. of days = x
21 × x = 14 × 39

Question 7.
Identify the type of proportion and fill in the blank boxes:

Solution:
Direct proportion
\(\frac{x}{y}\) = k = \(\frac{1}{20}\)
(i) x = 2; y = ?
\(\frac{2}{y}\) = \(\frac{1}{20}\) ⇒ y = 2 × 20 = 40

(ii) x = ?; y = 60
\(\frac{x}{60}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{60}{20}\) = 3

(iii) x = 4; y = ?
\(\frac{4}{y}\) = \(\frac{1}{20}\) ⇒ y = 80

(iv) x = 4; y = ?
\(\frac{8}{y}\) = \(\frac{1}{20}\) ⇒ y = 20 × 8 = 160

(v) x = ?; y = 180
\(\frac{x}{180}\) = \(\frac{1}{20}\)
x = \(\frac{180}{20}\) = 9

(vi) x = 12; y = ?
\(\frac{12}{y}\) = \(\frac{1}{20}\)
y = 12 × 20 = 240

(vii) x = ?; y = 360
\(\frac{x}{360}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{360}{20}\) = 18

(viii) x = 24; y = ?
\(\frac{24}{y}\) = \(\frac{1}{20}\) ⇒ y = 24 × 20 = 480

Question 8.
Identify the type of proportion and fill in the blank boxes:

Solution:
Inverse proportion
k = xy = 1 × 144 = 144
(i) x = 2; y = ?
2y = 144
y = 72

(ii) X = ?; y = 48
48x = 144
x = \(\frac{144}{48}\) = 3

(iii) x = 4; y = ?
4y = 144
y = \(\frac{144}{4}\) = 36

(iv) x = 8; y = ?
8 y = 144
y = \(\frac{144}{8}\) = 18

(v) x = ?; y = 16
16x = 144
y = \(\frac{144}{16}\) = 9

(vi) x = 12; y = ?
12y = 144
y = \(\frac{144}{12}\) = 12

(vii) x = ?; y = 9
9x = 144
x = \(\frac{144}{9}\) = 16

(viii) x = 24; y = ?
24y = 144
y = \(\frac{144}{24}\) = 6

Try These (Text book Page No. 38)

Question 1.
When x = 5 and y = 5 find k, if x and y vary directly.
Solution:
If x and y vary directly then \(\frac{x}{y}\) = k
Here x = 5; y = 5
∴ k = \(\frac{5}{5}\)
k = 1

Question 2.
When x and y vary inversely, find the constant of variation when x = 64 and y = 0.75
Solution:
Given
x =64, y = 0.75
and also given x and y vary inversely.
∴ xy = k. the constant of variation.
∴ Constant = 64 × 0.75
Constant of variation = 48

Think (Text book Page No. 38)

(i) When x and y are in direct proportion and if y is doubled, then what happens to x?
Solution:
If x and y are in direct proportion \(\frac{x}{y}\) = k, constant.
if y is doubled, then \(\frac{x}{2}\) must be equal to k. So x also to be doubled.

(ii) if \(\frac{x}{y-x}\) = \(\frac{6}{7}\) What is \(\frac{x}{y}\)?
Solution:
if \(\frac{x}{y-x}\) = \(\frac{6}{7}\)
\(\frac{y-x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) – \(\frac{x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + \(\frac{x}{x}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + 1
\(\frac{y}{x}\) = \(\frac{7+6}{6}\)
\(\frac{y}{x}\) = \(\frac{13}{6}\)
\(\frac{x}{y}\) = \(\frac{6}{13}\)

Try These (Text book Page No. 40)

Identify the different variations present in the following questions:
Question 1.
24 men can make 48 articles in 12 days. Then, 6 men can make …………. articles in 6 days.
Solution:
Let the required no. of articles be x

(i) Mens and days are Indirect variables.
(ii) Men and Articles are direct variables
(iii) Days and articles are also direct variables using formula.
Let P₁ = 24, D₁ = 12, W₁ = 48
P₂ = 6, D₂ = 6, W₂ = x

Question 2.
15 workers can lay a road of length 4 km In 4 hours. Then, …………. workers can lay a road of length 8 km in 8 hours.
Solution:
Let the required number of workers be x

Length and workers are direct variable as more length need more workers.
The proportion is 4 : 8 : : 15 : x ……….(1)
Hours and workers are indirect variables as more working hours need less men.
∴ The proportion is 8 : 4 : : 15 : x ………..(2)
Combining (1) and (2)

Product of the extremes = product of the means
4 × 8 × x = 8 × 4 × 15
x = \(\frac{8×4×15}{4×8}\)
x = 15 workers

Question 3.
25 women working 12 hours a day can complete a work in 36 days. Then, 20 women must ……….. work hours to complete the same work in 30 days.
Solution:
Let the required hours be x

As women increases hours to work decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{25}{20}\)
As days increases hours needed become less
∴ It is also an indirect variation.
∴ Multiplying factor is \(\frac{36}{30}\)
∴ x = 12 × \(\frac{25}{20}\) × \(\frac{36}{30}\)
x = 18 hours

Question 4.
In a camp, there are 420 kg of rice sufficient for 98 persons for 45 days. The number of days that 60 kg of rice will last for 42 persons is…………
Solution:
Let the required number of days be x.

If amount of rice is more it will last for more days.
∴ It is Direct Proportion.
∴ Multiplying factor is \(\frac{60}{420}\)
If men increases number of days the rice lasts decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{98}{42}\)
x = 45 × \(\frac{60}{420}\) × \(\frac{98}{42}\)

x = 15 days

Try These (Text book Page No. 44)

Question 1.
Vikram can do one-third of work in p days. He can do \(\frac{3}{4}\)th of work in ………… days.
Solution:
\(\frac{1}{3}\) of the work will be done in p days
∴ Full work will be completed in 3p days
\(\frac{3}{4}\)th of the work will be done in = 3p × \(\frac{3}{4}\) = \(\frac{9}{4}\)p = 2\(\frac{1}{4}\)p days

Question 2.
If m persons can complete a work in n days, then 4m persons can complete the same work in ……….. days and \(\frac{m}{4}\) persons can complete the same work in…….. days
Solution:
Given m persons complete a work in n days
(i) Then work measured in terms of Man days = mn
4 m men do the work it will be completed in \(\frac{mn}{4m}\) days = \(\frac{n}{4}\) days.
(ii) \(\frac{m}{4}\) persons can complete the same work in \(\frac{mn}{\frac{m}{4}}\) days = \(\frac{4mn}{m}\) = 4n days

Students can Download Science Term 3 Chapter 8 Conservation of Plants and Animals Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Science Solutions Term 3 Chapter 8 Conservation of Plants and Animals

Samacheer Kalvi 8th Science Conservation of Plants and Animals Text Book Exercise

I. Choose the best answer:

Question 1.
The plants found in a particular area are known as …………….
(a) fauna
(b) flora
(c) endemic
(d) rare
Answer:
(c) endemic

Question 2.
Deforestation means …………….
(a) cleaning of forest
(b) to grow plants
(c) to look after plants
(d) None of these
Answer:
(d) None of these

Question 3.
The Red Data Book gives a list of …………….
(a) endemic species
(b) extinct species
(c) natural species
(d) None of these
Answer:
(d) None of these

Question 4.
Insitu conservation is …………….
(a) off site conservation
(b) on site conservation
(c) Both a and b
(d) None of these
Answer:
(b) on site conservation

Question 5.
Wildlife Protection Act was implemented in …………….
(a) 1986
(b) 1972
(c) 1973
(d) 1971
Answer:
(b) 1972

II. Fill in the blanks:

1. WWF stands for …………….
2. The animal found in a particular area is known as …………….
3. Red Data Book is maintained by …………….
4. Mudhumalai Wildlife Sanctuary is located in ……………. district.
5. ……………. is observed as ‘World Wildlife day’.

Answer:

1. World Wildlife Fund
2. endemic
3. International Union for conservation of Nature
4. Nilgiris
5. March 3

III. Match the following:

1. Gir national park – Madhya Pradesh
2. Sundarabans National Park – Uttarangal
3. Indira Gandhi National Park – West Bengal
4. Corbett National Park – Gujarat
5. Kanha National Park – Tamil Nadu

Answer:

1. Gir national park – Gujarat
2. Sundarabans National Park – West Bengal
3. Indira Gandhi National Park – Tamil Nadu
4. Corbett National Park – Uttarangal
5. Kanha National Park – Madhya Pradesh

IV. Answer very briefly:

Question 1.
What is global warming?
Answer:
Gases methane and carbon dioxide accumulating in the atmosphere and trap the heat energy inside the atmosphere leading to increase in temperature is called Global warming.

Question 2.
What is known as extent species?
Answer:
Species which no longer exist on Earth are called extinct species.

Question 3.
Give few example for extinct species.
Answer:
Dinosaurs and Dodo are examples of extinct species.

Question 4.
Name two endangered animals.
Answer:

1. Snow Leo pard
2. Asiatic Lion

Question 5.
What is ICBN?
Answer:
International Code of Botanical Nomenclature. These are a set of International rules proposed by botanists to ensure a stable, universal and uniform system of naming plants.

V. Answer briefly:

Question 1.
What is biosphere reserve?
Answer:

1. Biosphere is a protected area where human population also forms the part of the system.
2. The area of these places will be around 5000 square kilometers. They conserve the eco system, species and genetic resources. These areas are set up mainly for economic development.

Question 2.
What is tissue culture?
Answer:
It is a technique of growing plant cells, tissues, organs, seeds or other plant parts in a sterile environment on a nutrient medium.

Question 3.
What is endangered species? Give two examples.
Answer:

1. An endangered species is an animal or a plant that is considered to be at the risk of extinction.
2. It means that there are only few of them left on the Earth and soon they might extinct.
3. Snow leopard, Bengal tiger, Asiatic lion, Purple frog and Indian giant squirrel are some of the endangered animals in India.

Question 4.
Write the advantages of the Red Data Book.
Answer:

1. It helps to evaluate the population of a particular species.
2. The data given in this book can be used to evaluate the species at the global level.
3. The risk of a species becoming globally extinct can be estimated with the help of this book.
4. It provides guidelines for implementing protective measures for endangered species.

Question 5.
Mention four main reasons for the conservation of forests.
Answer:

1. According to WWF (World Wildlife Fund) there has been 60% decrease in the size of population of animals, birds, fish, reptiles and amphibians over the past 40 years.
2. In order to leave something for the future generation, we need to conserve it now.
3. Conservation is the protection, preservation, management of wildlife and natural resource such as forest and water.
4. Conservation of biodiversity helps us to protect, maintain and recover endangered animals and plant species.

Question 6.
What do you understand by the term bio-magnification?
Answer:

1. Bio – magnification is the increase in contaminated substances due to the intoxicating environment.
2. The contaminants might be heavy metals such as mercury, arsenic, and pesticides such as polychlorinated biphenyls and DDT (Dichloro Diphenyl Trichloro ethane).
3. These substances are taken up by the organisms through the food they consume.
4. When the organisms in the higher food chain feed on the organisms in the lower food chain containing these toxins, these toxins get accumulated in the higher organisms.

Question 7.
What is PBR?
Answer:
1. Peoples Biodiversity Register is a document which contains comprehensive information on locally available bio-resources including landscape and demography of a particular area or village.

2. Bio-resources mean plants, animals and microorganisms or parts thereof, their genetic material and by-products with actual or potential use or value.

3. A Biodiversity Management Committee is set up in each local body which prepares the People’s Biodiversity Registers with the guidance and technical support of National Biodiversity Authority and the State Biodiversity Boards.

4. Preparation of this register promotes conservation, preservation of habitats and breed of animals and gathering of knowledge relating to biological diversity.

5. The register entails a complete documentation of biodiversity in the area related to the plant, food source, wildlife, medicinal source, traditional knowledge etc.

VI. Answer in detail:

Question 1.
What is Deforestation? Explain the causes and effects of Deforestation.
Answer:
Destruction of forests in order to make the land available for different uses is known as deforestation.
Causes of Deforestation:

• Fires and floods are the natural causes for deforestation.
• Human activities are responsible for deforestation include agricultural expansion, cattle breeding, illegal logging, mining, oil extraction, dam construction and infrastructure development.

1. Agricultural Expansion:

• With increasing population, there is an overgrowing demand for food production.
• Hence, large amount of trees are chopped down for crops and for cattle grazing.

2. Urbanization:

• Increase in population needs the expansion of cities.
• More land is needed to establish housing and settlement.
• Requirements like construction of roads, development of houses, mineral exploitation and expansion of industries also arise due to urbanisation.
• Forests are destroyed to meet all these needs.

3. Mining:

• Mining of coal, diamond and gold require a large amount of forest land.
• Large number of trees are cut down to clear the forest area. The waste that comes out from mining pollutes the environment and affects the nearby plants.

4. Construction of dams:
To provide water supply to the increasing population, large size dams are constructed. Hence, a great extend of forest area is being cleared.

5. Timber Production:

• Wood-based industries like paper, match-sticks, furniture need a substantial amount of wood supply.
• Wood is the most commonly used fuel, thus, a large number of trees are being cut down for fuel supplies.
• Illegal wood cutting is the main reason for the destruction of some valuable plants.

6. Forest fire:

• Forest fire be caused by humans, accidents or natural factors.
• Forest fires wipe out thousands of acres of forest land each year all over the world. This has tremendous effects on biodiversity and the economy as well.

7. Cyclones:
Cyclones destroy the trees on a massive scale.

Effects of Deforestation:
1. Extinction of species:
Deforestation has resulted in the loss of many wonderful species of plants and animals and many are on the verge of extinction.

2. Soil Erosion:

• When the trees are cut down, soils are exposed to the Sun’s heat.
• Extreme temperature of the summer dries up the moisture and makes the nutrients to evaporate, ft also affects the bacteria that helps in the breakdown of organic matter.

3. Water cycle:
When trees are cut down, the amount of water vapour released decreases for transpiration and hence there is a decrease in the rainfall. ‘

4. Floods:
When the trees are cut down, the flow of water is disrupted and it leads to flooding.

Question 2.
Discuss the advantages of Insitu and exsitu conservation.
Answer:
Advantages of In-situ conservation:

1. Species can be adapted to their habitat.
2. Species can interact with each other.
3. Natural habitat is maintained.
4. It is less expensive and easy to manage.
5. Interests of indigenous people are protected.

Advantages of Exquisite conservation:

1. It prevents the decline of species.
2. Endangered animals can be breeded in these ways.
3. Threatened species are breeded and released in natural environment.
4. It is useful for conducting research and scientific work.

Question 3.
Write about the types of conservation.
Answer:
Conservation is the protection, preservation, management of wildlife and natural resource such as forest and water.

Types:
In-situ conservation:
It is conservation of living resources within the natural ecosystem in which they occur.

1. National Parks:

• It is an area which is strictly reserved for the betterment of the wildlife.
• Here, activities like forestry, grazing or cultivation are not permitted.
• Eg – Guindy National Park in Chennai district.

2. Wildlife sanctuaries:

• A sanctuary is a protected area reserved for the conservation of animals only.
• Human activities like harvesting of timber, collection of forest products and private ownership rights are allowed here.
• Controlled interference like tourist activity is also allowed.

3. Biosphere reserves:

• It is a protected area where human population also forms the part of the system.
• The area of these places will be around 5000 square kilometers.
• They conserve the eco system, species and genetic resources.
• Eg – These areas are set up mainly for economic development.

Ex-situ Conservation:
It is the conservation of wildlife outside their habitat. Establishing zoos and botanical gardens, conservation of genes, seedling and tissue.

1. Botanical gardens:

• It is a place where flowers, fruits and vegetables are grown.
• These places provide a healthy and calm environment.

2. Zoological parks:

• Zoological parks are the areas where wild animals are conserved.
• In India there are about 800 zoological parks.

3. Tissue Culture:
It is a technique of growing plant cells, tissues, organs, seeds or other plant parts in a sterile environment on a nutrient medium.

4. Seed bank:
The seed bank preserves dried seeds by storing them in a very low temperature.

5. Cryo Bank:
It is a technique by which a seed or embryo is preserved at a very low temperature.

Question 4.
Write a note on Blue Cross.
Answer:
1. Blue Cross is a registered animal welfare charity in the United Kingdom, founded in 1897 as Our Dumb Friends League.

2. The vision of this charity is that every pet will enjoy a healthy life in a happy home.

3. The charity provides support for pet owners who cannot afford private veterinary treatment, helps to find homes for unwanted animals, and educates the public in the responsibilities of animal ownership.

4. Captain V. Sundaram founded the Blue Cross of India, the largest animal welfare organization of Asia in Chennai in the year 1959.

5. He was an Indian pilot and animal welfare activist. Now, Blue Cross of India the country’s largest animal welfare organizations and it runs several animal welfare events like pet adaptation and animal right awareness.

6. Blue Cross of India has received several international and national awards.

7. This organization is entirely looked after by volunteers.

8. The main office is located at Guindy, Chennai, with all amenities like hospitals, shelters, ambulance services and animal birth controls, etc.

9. Activities of the organization include, providing shelters, re-homing, adoption, animal birth control, maintaining hospitals and mobile dispensary and providing ambulance services.

VII. HOT:

Question 1.
Is it possible to find Dinosaurs today? Why?
Answer:

1. No. It is not possible to find Dinosaurs today since they have become extinct.
2. The reasons may be shortage of food, space or climate changes.

Question 2.
Animals are affected by Deforestation. How?
Answer:

1. Deforestation involves destruction of forests by man to make the lend available for different uses.
2. When forests are destroyed, the wild animals living there are left homeless.
3. They start moving out in search of shelter.
4. Since animals start moving out the food chains are also affected and animals cannot find food also.
5. They also get isolated from their groups and may be killed by accidents or hunted.

Question 3.
Why did the numbers of Tiger and Black buck decrease?
Answer:

1. The number of Black buck and Tigers have decreased due to excessive hunting, Deforestation and habitat degradation.
2. Tigers are also facing problems of prey depletion since the number of deers and other herbivores are decreasing due to disappearance of forests.

Samacheer Kalvi 8th Science Conservation of Plants and Animals Additional Question

I. Choose the correct answer:

Question 1.
…………… is not a green house gas.
(a) Oxygen
(b) Carbon dioxide
(c) Nitro-us oxide
(d) Methane
Answer:
(a) Oxygen

Question 2.
Chipko Movement was started in ……………
(a) 1980
(b) 1970
(c) 1960
(d) 1953
Answer:
(b) 1970

Question 3.
Each year …………… is celebrated as ‘World Biodiversity Day’.
(a) April 20
(b) May 22
(c) December 8
(d) October 12
Answer:
(b) May 22

Question 4.
…………… is not an endangered animal.
(a) Nilgiri Tahr
(b) Asiatic Lion
(c) Snow leopard
(d) Dodo duck
Answer:
(d) Dodo duck

Question 5.
Yeoman Butterfly has been declared as state butterfly of ……………
(a) Manipur
(b) Nagaland
(c) Tamil Nadu
(d) West Bengal
Answer:
(c) Tamil Nadu

II. Fill in the blanks:

1. The founder of Chipko movement was ……………
2. In a Cryo bank, the seeds are preserved in ……………
3. The variety of life forms is called …………….
4. Replanting of trees is called ……………
5. The details of endangered species can be viewed in ……………
6. Blue cross of India was established in …………… in India.
7. …………… has led to destruction of coral seeds.
8. …………… Biosphere reserve is located in Tamil Nadu.
9. National park is an example for …………… conservation.
10. World Wild life Day is celebrated on …………….

Answer:

1. Sunderlal Bahuguna
2. liquid nitrogen
3. Biodiversity
4. Reforestation
5. Red Data Book
6. Chennai
7. Biomagnification
8. Nilgiri
9. In-situ
10. March 3rd

III. Very short answer:

Question 1.
What is Reforestation?
Answer:
Reforestation is the natural or intentional replanting of the existing forests that have been destroyed through deforestation.

Question 2.
Differentiate Afforestation and Reforestation.
Answer:
Afforestation:

1. Trees are planted in new areas where there was no forest cover.
2. One sapling is planted to get one tree.
3. It is practiced to bring more area under forest

Reafforestation:

1. It is practiced in areas where forests have been destroyed.
2. Two saplings are planted to replace every felled tree
3. It is practiced to avoid deforestation.

Question 3.
Name two endangered plant species?
Answer:
Malabar Lily and Rafflesia flower.

Question 4.
What are the functions of CPCSEA?
Answer:

1. Approval of animal house facilities.
2. Permission for conducting experiments involving usage of animals.
3. CPCSEA stands for ‘The Committee for the Purpose of Control and Supervision of Experiments on Animals.
4. It is a statutory committee set up under the Preservation of Cruelty to Animals Act,1960.

Question 5.
The Red Data Book is maintained by which organisation?
Answer:

1. Red Data Book is maintained by the International Union for Conservation of Nature.
2. It is an international organization working in the field of nature conservation and sustainable use of natural resources.

IV. Long Answer:

Question 1.
List some measure to save endangered species.
Answer:

1. Animal species are endangered mainly because of hunting and poaching. If it is controlled, there can be a significant change in the number of endangered animals.
2. Controlling pollution can have a positive impact on animals, fish and birds all over the world.
3. By consuming less pollutants, we can protect the ecosystems.
4. Animals often mistake plastic for food and hence plastics harm and cause endangerment of many species. Limiting the amount of plastic and recycling it can save the endangered animals.
5. Recycling things and buying Eco friendly products will preserve the environment resources and hence the animals.
6. Pesticides and chemicals which cause damage to the environment should be avoided.
7. Planting native trees will provide food to the animals.

Question 2.
What are causes for endangerment of species?
Answer:

1. Loss of habitat.
2. Trees that provide food and shelter to so many species are destroyed due to human intervention.

Over hunting and poaching:
Large number of animals is hunted for their horns, skin, teeth and many other valuable products.

3. Pollution:

• Number of animals are affected by pollutions like air pollution and water pollution.
• In the recent years more number of animals is affected by wastes in the form of plastic.

4. New habitat:

• Sometimes animals are taken by people to new habitat where they do not naturally live.
• Some, of them may extinct and some may survive.
• The new ones may also get attacked by the species already living there and cause their extinction.

5. Chemicals:

• We use pesticides and other chemicals to get rid of damaging insects, pests or weeds.
• But they can also poison desired plants and animals if we do not use them correctly.

6. Diseases:
Diseases due to various unknown reasons may affect the animals and make them extinct.

7. Natural calamities:
Animals may also be destroyed due to natural disasters like flood and fire.

Question 3.
Explain the significance of Afforestation.
Answer:

1. Afforestation helps the wild animals and even humans to have shelter and to find their food source.
2. Through afforestation we can increase the supply of oxygen. Trees planted can increase the water vapour in the atmosphere to get the rainfall.
3. By planting trees the amount of carbon dioxide in the atmosphere can be reduced and thus the effects of air pollution, green house gases and global warming can be controlled.
4. Afforestation enables us to avoid desertification of land.
5. Barren lands experience strong winds and it causes soil erosion. Top soil is washed away during rainfall. Afforestation helps to grow more trees so that they can hold the top soil along with the nutrients.
6. Creating forests provides us fodder, fruits, firewood and many other resources.
7. Industries need specific type of trees. Afforestation helps us to grow a particular type of trees.

Students can Download Maths Chapter 1 Numbers Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Intext Questions

Exercise 1.1
Think (Text book page No. 3)

Question 1.
Is the square of a prime number, prime?
Solution:
No, the square of a prime number ‘P’ has at least 3 divisors 1, P and P². But a prime number is a number which has only two divisors, 1 and the number itself. So square of a prime number is not prime.

Question 2.
Solution:
The sum of two perfect squares, need not be always a perfect square. Also the difference of two perfect squares need not be always a perfect square. Bu the product of two perfect square is a perfect square.

Try These (Text book Page No. 3)

Consider the following square numbers:
(i) 441
(ii) 225
(iii) 289
(iv) 1089.
Express each of them as the sum of two consecutive positive integers.
Solution:
(i) 441 = 220 + 221
(ii) 225 = 112 + 113
(iii) 289 = 144 + 145
(iv) 1089 = 544 + 545

Think (Text book Page No. 4)

Question 1.
Take an even natural number, say, 46 (or any other even number of your choice). Try to express it as a sum of consecutive odd numbers starting with 1. Do you succeed?
Solution:
1 + 3 + 5 + 7 + 9 + 11 = 36
1 + 3 + 5 + 7 + 9 + 11 + 13 = 49
So 46 cannot be expressed as a sum of consecutive odd numbers starting with 1. But 36 also an even natural number. It can be expressed as the sum of consecutive odd numbers as 36 = 1 +3 + 5 + 7 + 9 + 11.

Question 2.
The square of an odd number can always be written as the sum of two consecutive natural numbers. Can the reverse statement be true? Is the sum of any two consecutive natural numbers a perfect square of a number?
Solution:
No, the reverse is not true. The sum of any two consecutive natural numbers need not be a perfect square of a number. Example: 35 + 36 = 71, not a perfect square.

Try These (Text book Page No. 4)

(i) Which among 256, 576, 960, 1025, 4096 are perfect square numbers?
(Hint: Try to extend the table of squares already seen).
Solution:
256 = 16²
576 = 24²
4096 = 64²
∴ 256, 576 and 4096 are perfect squares

(ii) One can judge just by look, that the each of the following numbers (82, 113, 2057, 24353, 8888,1972) is not a perfect square. Explain why?
Solution:
Because the unit digit of a perfect square will be 0, 1, 4, 5, 6, 9. But the given numbers have unit digits 2, 3, 7, 8. So they are not perfect squares.

(iii) Find the squares by diagonal method and also the ones digit in the squares of the following numbers: 11, 27, 42, 79, 146, 324, 520.
Solution:
Ones digit is the squares of

Think (Text book Page No. 5)

Consider the claim: “Between the squares of the consecutive numbers n and (n+1), there are 2n non-square numbers”. Can it be true? Find how many non-square numbers are there
(i) between 4 and 9?
(ii) between 49 and 64?
and verify the claim.
Solution:

Therefore we conclude that there are 2n non-square numbers between two consecutive square numbers.

Activity 1. (Text book Page No. 5)

Verify the following statements:
(i) The square of a natural number, other than 1, is either a multiple of 3 or exceeds a multiple of 3 by 1.
Verification

Hence verified that the square of a natural number other than 1, is either a multiple of 3 or exceeds a multiple of 3 by 1.

(ii) The square of a natural number, other than 1, is either a multiple of 4 or exceeds a multiple of 4 by 1.
Verification

So a perfect square leaves remainder 0 or 1 on division by 4.

(iii) The remainder of a perfect square when divided by 3, is either 0 or 1 but never 2.
Verification

So in all the cases the remainder is either 0 or 1, but not 2. Hence verified.

(iv) The remainder of a perfect square, when divided by 4, is either 0 or 1 but never 2 and 3.
Verification

From the table it is verified that the remainder of a perfect square when divided by 4 is either 0 or 1 but never 2 and 3.

(v) When a perfect square number is divided by 8, the remainder is either 0 or 1 or 4, but never be equal to 2, 3, 5, 6 or 7.
Verification

From the table it is verified when a perfect square number is divided by 8, the remainder is 0 or 1 or 4. but never be equal to 2, 3, 5, 6 or 7

Activity 2. (Text book Page No. 6)

Consider any natural number m > 1.We find that (2m, m² – 1, m² + 1) will form a Pythagorean triplet. (A little algebra can help you to verify this!). With this formula, generate a few Pythagorean triplets.
Solution:
For any natural number m, we have 2m, m² – 1, m² + 1 is a Pythagorean triplet.
(i) Consider 2m = 16 ⇒ m = 8
m² – 1 = 64 – 1 = 63
m² + 1 = 64 + 1 = 65
So Pythagorean triplet is 16, 63, 65.

(ii) Consider 2m = 18 ⇒ m = 9
m² – 1 = 81 – 1 = 80
m² + 1 = 81 + 1 = 82
So Pythagorean triplet is 18, 80, 82.

Try These (Text book Page No. 7)

Use the method of successive subtraction of odd numbers (starting from 1) to examine the following numbers and find if they are perfect squares or not. For perfect square numbers that you find, identify the square root.
(i) 144
(ii) 256
(iii) 360
Solution:
(i) 144
Step 1: 144 – 1 = 143
Step 2: 143 – 3 = 140
Step 3: 140 – 5 = 135
Step 4: 135 – 7 = 128
Step 5: 128 – 9 = 119
Step 6: 119 – 11 = 108
Step 7: 108 – 13 = 95
Step 8: 95 – 15 = 80
Step 9: 80 – 17 = 63
Step 10: 63 – 19 = 44
Step 11: 44 – 21 = 23
Step 12: 23 – 23 = 0.
We get 0 in the 12th step. ∴ 144 is a perfect square and ∴ \(\sqrt{144} \) = 12.

(ii) 256
Step 1: 256 – 1 = 255
Step 2: 255 – 3 = 252
Step 3: 252 – 5 = 247
Step 4: 247 – 7 = 240
Step 5: 240 – 9 = 231
Step 6: 231 – 11 = 220
Step 7: 220 – 13 = 207
Step 8: 207 – 15 = 192
Step 9: 192 – 17 = 175
Step 10: 175 – 19 = 156
Step 11: 156 – 21 = 135
Step 12: 135 – 23 = 112
Step 13: 112 – 25 = 87
Step 14: 87 – 27 = 60
Step 15: 60 – 29 = 31
Step 16: 31 – 31 = 0
We have subtracted odd numbers starting from 1 repeatedly from 256. We get 0 in the 1 step.
∴ 256 is a perfect square and \(\sqrt{256} \) = 16.

(iii) 360
Step 1: 360 – 1 = 359
Step 2: 359 – 3 = 356
Step 3: 356 – 5 = 351
Step 4: 351 – 7 = 344
Step 5: 344 – 9 = 335
Step 6: 335 – 11 = 324
Step 7: 324 – 13 = 311
Step 8: 311 – 15 = 296
Step 9: 296 – 17 = 279
Step 10: 279 – 19 = 260
Step 11: 260 – 21 = 239
Step 12: 239 – 23 = 216
Step 13: 216 – 25 = 191
Step 14: 191 – 27 = 164
Step 15: 164 – 29 = 135
Step 16: 135 – 31 = 104
Step 17: 104 – 33 = 71
Step 18: 71 – 35 = 36
Step 19: 36 – 37 = -1
We have subtracted successive odd numbers starting from 1, repeatedly from 360, we didn’t get zero.
∴ The given number is not perfect square.

Think (Text book Page No. 8)

In this case, if we want to find the smallest factor with which we can multiply or divide 108 to get a square number, what should we do?
Solution:
108 = 2 × 2 × 3 × 3 × 3 = 2² × 3² × 3
If we multiply the factors by 3, then we get
2² × 3² × 3 × 3 ⇒ 2² × 3² × 3² = (2 × 3 × 3)²
Which is a perfect square.
∴ Again if we divide by 3 then we get 2² × 3² ⇒ (2 × 3 )², a perfect square.
∴ We have to multiply or divide 108 by 3 to get a perfect square.

Exercise 1.2
Try These (Text book Page No. 11)

Find the square root by long division method.
(i) 400
(ii) 1764
(iii) 4356
Solution:

Try These (Text book Page No. 12)

Question 1.
Without calculating the square root, guess the number of digits in the square root of the following numbers:
(i) 14400
(ii) 100000000
(iii) 390625
Solution:
(i) \(\sqrt{14400}=\sqrt{144 \times 100}=\sqrt{144} \times \sqrt{100}\) = 12 x 10 = 120.
(ii) \(\sqrt{100000000}=\sqrt{10000 \times 10000}=\sqrt{10000} \times \sqrt{10000}\) = 100 × 100 = 10,000
(iii) \(\sqrt{390625}=\sqrt{25 \times 25 \times 25 \times 25}=\sqrt{25 \times 25} \times \sqrt{25 \times 25}\) = 25 x 25 = 625

Question 2.
Find the square root of
(i) 19.36
(ii) 1.2321
(iii) 116.64
Solution:

Think (Text book Page No. 13)

Fill up the table:

Solution:

Activity 3. (Text book Page No. 13)

Attempt to prepare a table of square root problems as in the above case to show that if a and b are two perfect square numbers, then \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) (b ≠ 0) We can use this idea to compute easily certain square-root problems.
Solution:

Try These (Text book Page No. 13)

Using the above method, find the square root of 1.2321 and 11.9025.
Solution:
(i) \(\sqrt{1.2321}=\sqrt{\frac{12321}{10000}}=\frac{111}{100}\) = 1.11
(ii) \(\sqrt{11.9025}=\frac{\sqrt{119025}}{\sqrt{10000}}=\frac{345}{100}\) = 3.45

Try These (Text book Page No. 14)

Put the numbers
(i) 4, \(\sqrt{14} \), 5 and
(ii) 7, \(\sqrt{65} \), 8 in ascending order.
Solution:
(i) 4, \(\sqrt{14} \), 5
Squaring all the numbers we get 4², \((\sqrt{14})^2 \) , 5² ⇒ 16, 14, 25
∴Ascending order: 14, 16, 25
Ascending order: \(\sqrt{14} \), 4, 5

(ii) 7, \(\sqrt{65} \), 8
Squaring 7, \(\sqrt{65} \) and 8 we get 7², \((\sqrt{65})^2 \), 8² ⇒ 49, 65, 64
Ascending order: 49, 64, 65
Ascending order: 7, 8, \(\sqrt{65} \)

Exercise 1.3
Try These (Text book Page No. 17)

Find the ones digit in the cubes of each of the following numbers,
(i) 17
(ii) 12
(iii) 38
(iv) 53
(v) 71
(vi) 84
Solution:
(i) 17
17 ends with 7, so its cube ends with 3. i.e, ones digit in 17³ is 3.
(ii) 12
12 ends with 2, so its cube ends with 8. i.e, ones digit in 12³ is 8.
(iii) 38
38 ends with 8, so its cube ends with 2. i.e, ones digit in 38³ is 2.
(iv) 53
53 ends with 3, so its cube ends with 7. i.e, ones digit in 53³ is 7.
(v) 71
71 ends with 1, so its cube ends with 1. i.e, ones digit in 71³ is 1
(vi) 84
84 ends with 4, so its cube ends with 4. i.e, ones digit in 84³ is 4.

Think (Text book Page No. 17)

Find the smallest number by which 675 must be divided to obtain a perfect cube.
Solution:

We find that 675 = 3 x 3 x 3 x 5 x 5
= 3 x 3 x 3 x 5 x 5
Here we have an ungrouped 5 x 5.
So if we divide 675 by 5 x 5 then the new number will be a perfect cube.
ie. 675 ÷ 25 = 27
The new number is 27
\(\sqrt[3]{27}=\sqrt[3]{3 \times 3 \times 3}\) = 3

Exercise 1.4
Try These (Text book Page No. 21)

Expand the following numbers using exponents:
(i) 8120
(ii) 20305
(iii) 3652.01
(iv) 9426.521
Solution:
(i) 8120 = (8 x 1000) + (1 x 100) + (2 x 10) + 0 x 1
= (8 x 10³) + (1 x 10²) + (2 x 10¹)

(ii) 20305 = (2 x 10000) + (0 x 1000) + (3 x 100) + (0 x 10) + (5 x 1)
= (2 x 10⁴) + (3 x 10²) + 5

(iii) 3652.01 = 3000 + 600 + 50 + 2 + \(\frac{0}{10}+\frac{1}{100}\)
= (3 x 1000) +(6 x 100) + (5 x 10) + (2 x 1) + (1 x \(\frac{1}{100}\))
= (3 x 10³) + (6 x 10²) + (5 x 10¹) + 2 + (1 x 10^-2)

(iv) 9426.521 = (9 x 1000) + (4 x 100) + (2 x 10) + (6 x 1) + \(\left(\frac{5}{10}\right)+\left(\frac{2}{100}\right)+\left(\frac{1}{1000}\right)\)
= (9 x 10³) + (4 x 10²) + (2 x 10¹) + 6 + (5 x 10^-1) + (2 x 10^-2) + (1 x 10^-3)

Try These (Text book Page No. 23)

Verify the following laws (as we did above). Here, a, b are non-zero integers and m, n are any integers.
1. Zero exponent rule: a⁰ = 1.
2. Product of same powers to power of product rule: a^m x b^m = (ab)^m
3. Quotient of same powers to power of quotient rule: \(\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}\)
Verification:
Let a = 2; b = 3; m = 2
1. a⁰ = 2⁰ = 1.
2. a^m x b^m = 2² x 3² = 4 x 9 = 36 = (2 x 3)²
3. \(\frac{a^{m}}{b^{m}}=\frac{2^{2}}{3^{2}}=\frac{4}{9}=\left(\frac{2}{3}\right)^{2}\)

Try These (Text book Page No. 26)

Question 1.
Write in standard form: Mass of planet Uranus = 8.68 x 10²⁵ kg.
Solution:
Mass of Planet Uranus = 86800000000000000000000000 kg
[23 zeros after 868]

Question 2.
Write in scientific form:
(i) 0.000012005
(ii) 4312.345
(iii) 0.10524
(iv) The distance between the Sun and the planet Saturn 1.4335 x 10¹² miles.
Solution:

(iv) The distance between Sun and the planet Saturn is 1.4335 x 10¹² miles

Activity 4. (Text book Page No. 29)

Observe that
2³ – 1³ = 1 + 2 x 1 x 3
3³ – 2³ = 1 + 3 x 2 x 3
4³ – 3³ = 1 + 4 x 3 x 3
Find the value of 15³ – 14³ in the above pattern.
Solution:
15³ – 14³ = 1 + 15 x 14 x 3

Think (Text book Page No. 29)

1³ = 1 = 1
2³ = 8 = 3 + 5
3³ = 27 = 7 + 9 + 11
Observe and continue this pattern to find the value of 73 as the sum of consecutive odd numbers.
1³ = 1 = 1
2³ = 8 = 3 + 5
3³ = 27 = 7 + 9 + 11
4³ = 64 = 13 + 15 + 17 + 19
5³ = 125 = 21 + 23 + 25 + 27 + 29
6³ = 216 = 31 + 33 + 35 + 37 + 39 + 41
7³ = 343 = 43 + 45 + 47 + 49 + 51 + 53 + 55

Students can Download Science Term 3 Chapter 6 Chemistry in Everyday Life Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Science Solutions Term 3 Chapter 6 Chemistry in Everyday Life

Samacheer Kalvi 8th Science Chemistry in Everyday Life Text Book Exercise

I. Choose the best answer:

Question 1.
The chemical mixed with LPG that helps in the detection of its leakage is …………………
(a) methanol
(b) ethanol
(c) camphor
(d) mercapton
Answer:
(d) mercapton

Question 2.
Which is known as syn gas?
(a) Marsh gas
(b) Water gas
(c) Producer gas
(d) Coal gas
Answer:
(b) Water gas

Question 3.
The unit of calorific value of fuel is …………………
(a) kilo joule per mole
(b) kilo joule per gram
(c) kilo joule per kilo gram
(d) joule per kilo gram
Answer:
(c) kilo joule per kilo gram

Question 4.
………………… is the coal of superior quality.
(a) Peat
(b) Lignite
(c) Bituminous
(d) Anthracite
Answer:
(d) Anthracite

Question 5.
The main component of natural gas is …………………
(a) methane
(b) ethane
(c) propane
(d) butane
Answer:
(a) methane

II. Fill in the blanks:

1. Producer gas is a mixture of …………..
2. ………….. is known as marsh gas.
3. The term petroleum means …………..
4. Heating coal in the absence of air is called ……………
5. An example for fossil fuel is …………..

Answer:

1. Carbon monoxide and nitrogen
2. Methane
3. Rock oil
4. Destructive distillation
5. Coal

III. Match the following:

Question 1.

1. Octane rating – (a) Diesel
2. Cetane rating – (b) Methane
3. Simplest hydrocarbon – (c) Petrol
4. Peat – (d) Bown in colour
5. Lignite – (e) First stage coal

Answer:

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

IV. Answer briefly:

Question 1.
What do you mean by catenation?
Answer:
The property of carbon atom to form bonds with itself resulting in a single large structure or chain is called catenation.

Question 2.
Mention the advantages of natural gas.
Answer:

1. It produces lot of heat as it is easily burnt.
2. It does not leave any residue.
3. It bums without smoke and so causes no pollution.
4. This can be easily supplied through pipes.
5. It can be directly used as fuel in homes and industries.

Question 3.
Expand CNG. List out its uses.
Answer:
CNG – Compressed Natural Gas.

1. It is the cheapest and cleanest fuel.
2. Vehicles using this gas produce less carbon dioxide and hydrocarbon emission.
3. It is less expensive than petrol and diesel.

Question 4.
Identify the gas known as syngas. Why is it called so?
Answer:
Water Gas is also called as syngas or synthesis gas as it is used to synthesize methanol and simple hydrocarbons. It is used as an industrial fuel also.

Question 5.
Anthracite is known as the highest grade coal. Give reason.
Answer:

1. Anthracite is the highest grade coal.
2. It has a very light weight and the highest heat content.
3. Anthracite coal is very hard, deep black and shiny.
4. It contains 86-97% carbon and has a heating value slightly higher than bituminous coal.
5. It bums longer with more heat and less dust.

Question 6.
Distinguish between octane number and cetane number.
Answer:
Octane Number:

1. Octane rating is used for petrol
2. It measures the amount of octane present in petrol
3. Octane number of petrol can be increased by adding benzene or toluene.
4. The fuel with high octane number has low cetane number

Cetane Number:

1. Cetane rating is used for diesel
2. It measures the ignition delay of the fuel in diesel engine.
3. Cetane number of diesel can be increased by adding acetone.
4. The fuel with high cetane number has low octane number

Question 7.
Name the places in Tamilnadu harnessing wind energy from wind mills.
Answer:
Wind mills are mostly located at Kayathar, Aralvaimozhi, Palladam and Kudimangalam in Tamil Nadu.

Question 8.
Solar energy is a non – depleting energy. Justify.
Answer:

1. Solar energy is the only viable fuel source of non – depleting nature for, Sun provides a free and renewable source of energy.
2. It is the renewable type of energy without endangering the environment.
3. It is the potential source to replace the fossil fuel in order to meet the needs of the world. With the advancements in science and technology, solar energy has become more affordable, and it can overcome energy crisis.

V. Answer in detail:

Question 1.
Explain the different types of coal.
Answer:
Lignite:

1. Lignite is a brown coloured coal of lowest grade.
2. It has least content of carbon. The carbon content of lignite is 25 – 35%.
3. Lignite contains a high amount of water and makes up almost half of our total coal reserves.
4. It is used for electricity generation.
5. It is used to generate synthetic natural gas and produce fertilizer products.

Sub-bituminous:

1. When lignite becomes darker and harder over time, sub-bituminous coal is formed.
2. Sub-bituminous coal is a black and dull coal.
3. It has higher heating value than lignite and contains 35-44% carbon.
4. It is used primarily as fuel for electricity power generation.
5. This coal has lower sulfur content than other types and bums cleaner.

Bituminous:

1. With more chemical and physical changes, sub-bituminous coal is developed into bituminous coal.
2. It is dark and hard. It contains 45-86% carbon. It has high heating value.
3. It is used to generate electricity.
4. Other important use of this coal is to provide coke to iron and steel industries.
5. By-products of this coal can be converted into different chemicals which are used to make paint, nylon and many other items.

Anthracite:

1. It is the highest grade coal. It is hard and dark black in colour.
2. It has a very light weight and the highest heat content.
3. Anthracite coal is very hard, deep black and shiny.
4. It contains 86-97% carbon and has a heating value slightly higher than bituminous coal.
5. It bums longer with more heat and less dust.

Question 2.
What is known as destructive distillation? Write about the products obtained from petroleum.
Answer:
Coal when heated in the absence of air does not bum but produces many by-products. This process of heating coal in the absence of air is called destructive distillation of coal.
Products obtained from petroleum:

1. Liquefied Petroleum Gas or LPG
2. Diesel and petrol
3. Kerosene
4. Lubricating oil
5. Paraffin
6. Bitumen or asphalt
7. Refinery Gas
8. Naphtha
9. Fuel Oil
10. Chemicals
11. Jet fuel
12. Waxes
13. Polishes

Question 3.
What are the different types of fuels?
Answer:
Fuels are classified into solid, liquid and gaseous fuels according to their physical state.
Solid fuels:

1. Fuels like wood and coal are in solid state and they are called solid fuels.
2. This type of fuel was the first one to be used by man.
3. These fuels are easy to store and transport.
4. The production cost is also very low.

Liquid fuels:

1. Most of the liquid fuels are derived from the fossil remains of dead plants and animals petroleum oil, coal tar and alcohol are some of the liquid fuels.
2. These fuels give more energy on burning and bum without ash.

Gaseous fuel:

1. Coal gas, oil gas, producer gas and hydrogen are some of the gaseous fuels.
2. It can be easily transported through pipes and they do not produce pollution.

Samacheer Kalvi 8th Science Chemistry in Everyday Life Additional Questions

I. Choose the correct answer:

Question 1.
Dead and decaying plants and animals release ………………… gas.
(a) propane
(b) butane
(c) methane
(d) pentane
Answer:
(c) methane

Question 2.
………………… is an odourless and highly inflammable gas.
(a) Propane
(b) Butane
(c) Ethane
(d) Methane
Answer:
(a) Propane

Question 3.
Which gas is used as fuels and solvents in the laboratory?
(a) Natural gas
(b) CNG
(c) Methane
(d) Butane
Answer:
(d) Butanel

Question 4.
Which of the following is referred as wood gas?
(a) Water gas
(b) Producer gas
(c) Coal gas
(d) None
Answer:
(b) Producer gas

Question 5.
Which is used in museums to protect the monuments?
(a) CNG
(b) Producer Gas
(c) Water gas
(d) Natural gas
Answer:
(d) Natural gas

Question 6.
………………… is a mixture of methane and carbon dioxide.
(a) Bio-gas
(b) Water gas
(c) Coal gas
(d) None
Answer:
(a) Bio-gas

Question 7.
………………… is a gaseous mixture of carbon monoxide and hydrogen.
(a) Bio-gas
(b) Natural gas
(c) Water gas
(d) Producer gas
Answer:
(b) Natural gas

Question 8.
The breaking down of organic matter in an anaerobic condition leads to the formation of ………………….
(a) coal gas
(b) natural gas
(c) bio-gas
(d) CNG
Answer:
(c) Bio-gas

Question 9.
When lignite becomes darker and harder over time ………………… coal is formed.
(a) lignite
(b) bituminous
(c) anthracite
(d) sub-bituminous
Answer:
(d) sub-bituminous

Question 10.
Which one among the following is an extremely strong but lightweight material is used in mountain bikes and tennis rackets?
(a) Cotton fibre
(b) Carbon fibre
(c) Plastic fibre
(d) None
Answer:
(b) Carbon fibre

II. Fill in the blanks:

1. Hydrogen gas obtained from natural gas is used in the production of …………….
2. Hydrocarbons are present in different ……………. and …………….
3. Hydrocarbons are less dense than …………….
4. Sewage sludge can also be decomposed by microorganisms to produce methane gas along with impurities like carbon dioxide and ……………..
5. ……………. can also be used as refrigerants.
6. ……………. is used as a fuel gas and propellant in aerosol sprays such as deodorants.
7. ……………. natural gas can be extracted through drilling wells.
8. ……………. coal is a black and dull coal.
9. ……………. is used to make face packs and cosmetics.
10. Activated carbon used in filters for water and air purification and in ……………. machines is obtained from coal.

Answer:

1. fertilizers
2. trees, plants
3. water
4. hydrogen sulphide
5. Propane
6. Butane
7. Conventional
8. Sub-bituminous
9. Activated charcoal
10. kidney dialysis

III. Write true or false- if false, write the correct statement:

Question 1.
Coal helps to create alumina refineries.
Answer:
True

Question 2.
Coke is used in the preparation of dyes, explosives, paints synthetic fibers and pesticides.
Answer:
False
Correct statement:
Coal tar is used in the preparation of dyes, explosives, paints, synthetic fibers and pesticides.

Question 3.
Coal is also known as black diamond owing to its precious nature.
Answer:
True

Question 4.
Crude oil is used as a sealant for waterproofing various surfaces.
Answer:
True

Question 5.
The process of separating petroleum into useful by-products and removal of undesirable impurities is called distillation.
Answer:
False
Correct statement:
The process of separating petroleum into useful by-products and removal of undesirable impurities is called refining.

IV. Match the following:

Question 1.

1. Hydrocarbons – (a) Marsh gas
2. Methane – (b) Cleanest fuel
3. Butane – (c) Catenation
4. CNG – (d) Polystryrene

Answer:

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

V. Short answer questions:

Question 1.
Write any 2 properties of hydrocarbons.
Answer:

1. Most of the hydrocarbons are insoluble in water.
2. Hydrocarbons are less dense than water. So they float on top of water.
3. Most hydrocarbons react with oxygen to produce carbon dioxide and water.

Question 2.
Write the uses of natural gas.
Answer:

1. Natural gas is used as an industrial and domestic fuel.
2. It is used in thermal power stations.
3. It is used as fuel in vehicles as an alternative for petrol and diesel.

Question 3.
Write a note on bio-gas.
Answer:

1. Bio-gas is a mixture of methane and carbon dioxide.
2. It is produced by the decomposition of plant and animal waste which form the organic matter.
3. The breaking down of organic matter in anaerobic condition (ie., in the absence of oxygen) leads to the formation of biogas.
4. It is an example for renewable source of energy.

Question 4.
What is surface mining?
Answer:
If the coal beds lie within 22 feet of the Earth’s surface, the top soil is removed and coal is dug out. This is called surface mining.

Question 5.
What is deep mining?
Answer:
In some places, coal beds are found very deep inside the Earth. In that case, underground tunnels are made to get this coal. This is called underground mining or deep mining.

Question 6.
Write a note on coke.
Answer:

1. Coke contains 98% carbon. It is porous, black and the purest form of coal.
2. It is a good fuel and burns without smoke.
3. It is largely used as a reducing agent in the extraction of metals from their ores.
4. It is also used in making fuel gases like producer gas and water gas which is a mixture of carbon monoxide and hydrogen.

Question 7.
What are the characteristics of fuel?
Answer:

1. It should be readily available.
2. It should be easily transportable
3. It should be less expensive
4. It should have high calorific value
5. It should produce large amount of heat
6. It should not leave behind any undesirable substances.

Question 8.
Write a note on bio – diesel.
Answer:
Bio diesel is a fuel obtained from vegetable oils such as soya bean oil, jatropha oil, corn oil, sunflower oil, cotton seed oil, rice-bran oil and rubber seed oil.

VI. Long Answer type Questions:

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

1. Coal is used to generate heat and electricity.
2. It is used to make derivatives of silicon which are used to make lubricants, water repellents, resins, cosmetics, hair shampoos and toothpaste.
3. Activated charcoal is used to make facepacks and cosmetics.
4. It is used to make paper.
5. It helps to create alumina refineries.
6. Carbon fibre is an extremely strong but lightweight material is used in construction, mountain bikes, and tennis rackets.
7. Activated carbon used in filters for water and air purification and in kidney dialysis machines is obtained from coal.

Question 2.
Explain destructive distillation of Coal.
Answer:

1. The destructive distillation of coal can be carried out in the laboratories.
2. Finely powdered coal is taken in a test tube and heated.
   • At a particular temperature, coal breaks down to produce coke, coal tar, ammonia and coal gas.
   • Coal tar is deposited at the bottom of the second test tube and coal gas escapes out through the side tube.
   • Ammonia produced is absorbed in the water, forming ammonium hydroxide.
   • Finally, a black residue called coke is left in the first tube.

Students can Download English Poem 2 Lesson in Life Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 8th English Book Solutions Guide Pdf  helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th English Solutions Term 1 Poem Chapter 2 Lesson in Life

A. Comprehension Questions.

Question 1.
What is planting a flower is compared to?
Answer:
Planting a flower is compared to the blossoming of a friendship.

Question 2.
What does the tiniest creature need?
Answer:
Even the tiniest creature needs space in the world.

Question 3.
What do the smallest gifts deserve?
Answer:
Even the smallest gifts deserve a warm thank you.

Question 4.
What will happen if you fail to give importance to others?
Answer:
If we fail to give importance to others, they will feel sad.

Question 5.
What do you learn from your lessons in life?
Answer:
I express my friendship with others by respecting them.

Question 6.
Pick and write the rhyming words from the third stanza.
Answer:
‘sad’ and ‘bad’ are the rhyming words.

Question 7.
“Having a friend is like planting a flower”
What is the figure of speech used in this line?
Answer:
The figure of speech used in this line is “simile”.

Figures of speech.

a. Simile: Answer the following:

Question 1.
Write a sentence using ‘as fast as the wind’.
Answer:
He runs as fast as the wind.

Question 2.
Write a simile using the word ‘like’.
Answer:
They fought like cats and dogs.

Question 3.
Create a simile using the word ‘as’.
Answer:
As sweet as honey.

Question 4.
What does ‘smart as a fox’ mean?
Answer:
It means to be clever or cunning as a fox.

b. Metaphor:

Question 1.
Which of the given options is a Metaphor?
(a) Life is like a chocolate box.
(b) Raj is like his twin brother
(c) His words are pearls of wisdom
(d) The bus is slow as a snail.
Answer:
(c) His words are pearls of wisdom.

Question 2.
What does ‘The world is a stage” mean?
Answer:
The world is like a stage where all of us are merely actors.

Question 3.
Identify the Metaphor in the sentence:
Answer:
Her hair is always a rat’s nest in the morning, a Hair is always a rat’s nest.

Question 4.
Write a sentence on your own that includes a Metaphor.
Answer:
He was a lion in the battle.

Lesson in Life Additional Questions

I. Poem Comprehension.

Question 1.
Let’s be aware as we walk on this planet
Even the tiniest creature needs room.
(a) What should we aware of ?
Answer:
We should be aware that even the tiniest creature needs room.
(b) What do you mean by the term ‘as we walk on this planet’?
Answer:
It means that as long ‘as we live in this world’.

Question 2.
Lessons in life aren’t always so simple Nothing you’re given will ever come free
(a) Are ‘lessons in life’ simple?
Answer:
No, lessons in life are not so simple.
(b) What does the second line mean?
Answer:
It means that you are not going to get anything free of cost.

Question 3.
Even the smallest of gifts deserves “thank you”.
What should you say, even when receive a small gift?
Answer:
We should say ‘thank you’.

Question 4.
Know that you matter and you make a difference
(a) What does the above line tell us?
Answer:
The above line tells us that every human being is unique and important

II. Poetic Devices.

Question 1.
Having a friend is like planting a flower
(a) What is the poetic device used here?
Answer:
Simile is used here.

Question 2.
Lessons in life aren’t always so simple
(a) Pick out the alliterated words.
Answer:
Lessons – life; so – simple are the alliterated words.

Question 3.
Having a friend is like planting a flower
Show love and kindness it one day will bloom
Let’s be aware as we walk on this planet
Even the tiniest creature needs room.
(a) Mention the rhyme scheme in this stanza.
Answer:
‘abcb’ is the rhyme scheme in this stanza.

III. Short Questions and Answers.

Question 1.
How can having a friend be like planting a flower?
Answer:
Both friendship and planting flowers will bloom one day, when we show love, kindness and care every day.

Question 2.
When do we feel sad?
Answer:
We feel sad when people don’t remember us.

Question 3.
How does the poet explain that everyone is important?
Answer:
The poet says that everyone is important and each makes a difference. Since we are important we must not allow others to shame us or make us feel bad.

Question 4.
Why do you think the second stanza has been repeated as the 4 stanza?
Answer:
The message that we learn a lot of lessons in life after going through difficulty and that if anyone gives us even a very small present, it is not without effort, the importance of gratitude in life and respect for people, have been repeated.

Question 5.
To whom does T and’You’refer to?
Answer:
I refers to the poet and you refers to the reader.

IV. Paragraph Questions with Answers.

Question 1.
How should we treat a flowering plant and true friendship?
Answer:
The poet compares planting flowers to maintaining friendship, because both need care, kindness and love to blossom. We must understand that as we move on this planet, every living thing, even the smallest needs its own space. Nothing is free of cost, either hard work, a lot of thought or money are involved in the gift. So we must learn to thank the person, who gives a gift and respect human beings.

Question 2.
What is the importance of human beings on earth?
Answer:
Every human being is unique. He/she deserves to be respected. When one is forgotten by the other, it makes him feel very sad. Every person makes a difference in life and his presence is needed. We must never put anyone to shame and no one has the authority to make others feel bad about themselves. We must respect and treat everyone well and be considerate towards others.

Lesson in Life Summary

In the poem Lessons in life the poets Daniel Ho and Bridgette Bryant bring out the importance of friendship in life and they add value to it by comparing it to the episode of planting flowers.

Just like making a flower bloom, friendship also blossoms only when there is constant care, love and kindness showered on them for a long period of time. Lessons that are taught by life come out of facing great difficulty. Similarly every small gift has a great deal of effort behind it. One should respect it and never forget to say ‘thank you’. We feel sad when we are forgotten by others and similarly others feel sad when they are forgotten. Everyone is unique and no one must be put to shame.

The poets repeat the values of gratitude, love and respect in the final stanza to make everyone understand that without these, no one can live happily. Hence happy friendship blossoms and planting flowers bloom only with love and kindness reciprocated.