Category: Class 10

Students can download Maths Chapter 8 Statistics and Probability Ex 8.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.4

Question 1.
If P (A) = \(\frac{2}{3}\), P(B) = \(\frac{2}{5}\), P(A ∪ B) = \(\frac{1}{3}\), then find P(A ∩ B).
Answer:
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)
\(\frac{1}{3}=\frac{2}{3}+\frac{2}{5}\) – P (A ∩ B)

Question 2.
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A∩B)=016. Find (i) P(not A)
(ii) P(not B)
(iii) P(A or B)
Solution:
(a) P(A) = 0.42 ;
P(B) = 0.48
P(A∩B) = 0.16
(i) P(not A) = P(\(\overline{\mathbf{A}}\)) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = P(\(\overline{\mathbf{B}}\)) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.42 + 0.48 – 0.16
= 0.74

Question 3.
If A and B are two mutually exclusive events of a random experiment and P (not A) = 0.45, P (A ∪ B) = 0.65, then find P(B).
Answer:
P(not A) = 0.45
1 – P (A) = 0.45
P (A) = 1 – 0.45 = 0.55
P(A ∪ B) = P (A) + P (B)
0. 65 = 0.55 + P(B)
0. 65 – 0.55 = P(B)
0.10 = P (B)
P(B) = 0.1

Question 4.
The probability that atleast one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(\(\overline{\mathbf{A}}\)) + P(\(\overline{\mathbf{B}}\)).
Solution:
P(A∪B) = 0.6
P(A∩B) = 0.2
P(A) + P(B) = [1 – P(A∪B)] + [1 – P(A∩B)] = [1 – 0.6] + [1 – 0.2]
= 0.4 + 0.8 = 1.2

Question 5.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B happen.
Answer:
Here P(A) = 0.5, P (B) = 0.3
P(A ∪ B) = P (A) + P(B) [A and B are mutually exclusive]
= 0.5 + 0.3
= 0.8
Probability of neither A nor [P(A ∪ B)’] = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Question 6.
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (S) = 36
Let A be the event of getting an even number on the first time
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (A) = 18
\(P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}\)
(ii) Let B be the event of getting a total of face sum 8.
B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{36}\)
A ∩ B = {(2, 6) (4, 4) (6, 2)}
n(A ∩ B) = 3
P(A ∩ B) = \(\frac{3}{36}\)
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)

The required probability = \(\frac{5}{9}\)

Question 7.
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being either a red king or a black queen.
Answer:
n(S) = 52
Let A be the event of getting a red king
n(A) = 2
\(P(A)=\frac{n(A)}{n(S)}=\frac{2}{52}\)
Let B be the event of getting a black Queen king
n(B) = 2
\(P^{\prime}(B)=\frac{n(B)}{n(S)}=\frac{2}{52}\)
It A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
\(=\frac{2}{52}+\frac{2}{52}=\frac{4}{52}=\frac{1}{13}\)
The required probability is \(\frac{1}{13}\)

Question 8.
A box contains cards numbered 3, 5, 7, 9,… 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
Answer:
Sample space = {3, 5, 7, 9,…….,35, 37}
n(S) = 18
Let A be the event of getting a multiple of 7
A = {7, 21, 35}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{18}\)
Let B be the event of getting a prime number
B = {3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37}
n(B) = 11

Probability of getting a multiple of 7 or a prime number = \(\frac{13}{18}\)

Question 9.
Three unbiased coins are tossed once. Find the probability of getting atmost 2 tails or atleast 2 heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting atmost 2 tails.
A = {HTT, THT, TTH, HHT, HTH, THH, HHH}
n(A) = 7

Probability of getting atmost two tails or atleast 2 heads = \(\frac{7}{8}\)

Question 10.
The probability that a person will get an electrification contract is \(\frac{3}{5}\) and the probability that he will not get plumbing contract is \(\frac{5}{8}\). The probability of getting atleast one contract is \(\frac{5}{7}\). What is the probability that he will get both?
Answer:
Let A and B represent the event of getting electrification control and plumbing contract.


Probability of getting both the job is \(\frac{73}{280}\)

Question 11.
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?
Answer:
Total number of people in a town is 8000.
n(S) = 8000
Total number of females = 3000
Let A be the event of getting number of females
n(A) = 3000
\(P(A)=\frac{n(A)}{n(S)}=\frac{3000}{8000}\)
Number of people over 50 years = 1300
Let B be the event of getting number of people over 50 years.
n(B) = 1300
\(P(B)=\frac{n(B)}{n(S)}=\frac{1300}{8000}\)
Given 30% of the females are over 50 years.


Proability of getting either a female or over 50 years = \(\frac{17}{40}\)

Question 12.
A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}\)
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}\)
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3



The probability is 1.

Question 13.
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P (A ∩ B) = \(\frac{1}{6}\), P(B ∩ C) = \(\frac{1}{4}\), P(A ∩ C) = \(\frac{1}{8}\), P(A ∪ B ∪ C) = \(\frac{9}{10}\) and P (A ∩ B ∩ C) = \(\frac{1}{15}\), then find P(A), P(B) and P(C)?
Answer:
By the given condition,
P(B) = 2 P(A), P(C) = 3 P(A)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

Question 14.
In a class of 35, students are numbered from 1 to 35. The ratio of boys and girls is 4 : 3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.
Answer:
Sample space (S) = {1, 2, 3,… ,35}
n(S) = 35
Total number of students = 35
Number of boys = \(\frac{4}{7}\) × 35 = 20 [Boys Numbers = {1, 2, 3,…, 20}]
Number of girls = \(\frac{3}{7}\) × 35 = 15 [Girls Numbers = { 21, 22,…, 35}]
Let A be the event of getting a boy role number with prime number
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\frac{8}{35}\)
Let B be the event of getting girls roll number with composite number.
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}
n(B) = 12
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}\) = \(\frac{12}{35}\)
Let C be the event of getting an even roll number.
C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}
n(C) = 17


Probability of getting roll number is \(\frac{29}{35}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.3

Question 1.
Write the sample space for tossing three coins using tree diagram.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Question 3.
If A is an event of a random experiment such that P(A) : P(\(\bar{A}\)) = 17 : 15 and n(s) = 640 then find (i) P(\(\bar{A}\))
(ii) n(A)
Answer:

Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting consecutive tails
A = {HTT, TTH, TTT}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{8}\)

Question 5.
At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Solution:
22² = 484
31² = 961
23² = 529
32² = 1024
23, 24, 25, 26, 27, 28, 29, 30, 31 has squares below 500 × 1000.
(i) P(first player wins a prize) = \(\frac{9}{1000}\)
(ii) P(second player ins if first has won) = \(\frac{8}{999}\)

Question 6.
A bag contains 12 blue balls and x red balls. If one ball is drawn at random (i) what is the probability that it will be a red ball? (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Answer:
Sample space = 12 + x
n(S) = x + 12
(i) Let A be the event of getting a red ball
n(A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\left(\frac{x}{x+12}\right)\)
(ii) 8 more red balls are added
Sample space = x + 12 + 8 = x + 20
Number of red balls = x + 8
Probability of drawing red ball = \(\frac{x+8}{x+20}\)
By the given condition
\(\frac{x+8}{x+20}=2\left(\frac{x}{x+12}\right)\)
(x + 8)(x + 12) = 2x(x + 20)
x² + 20x + 96 = 2x² + 40x
x² + 20x – 96 = 0
(x + 24)(x – 4) = 0
x = -24 (or) x = 4
The value of x = 4 (Number of balls will not be negative)
The probability of getting red balls = \(\left(\frac{x}{x+12}\right)=\frac{4}{16}=\frac{1}{4}\)

Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Answer:
(i) Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A be the event of getting doublet
A = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting a product is a prime number.
B = {(1, 2) (1, 3) (1, 5) (2, 1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting a sum is a prime number
C = {(1, 1) (1, 2) (1, 4) (1, 6) (2, 1) (2, 3) (2, 5) (3, 2), (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)}
n(C) = 15
\(P(C)=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting a sum is 1
n(D) = 0
\(P(D)=\frac{n(D)}{n(S)}=\frac{0}{36}=0\)
Probability of getting a sum is 1 is 0

Question 8.
Three fair coins are tossed together. Find the probability of getting
(i) all heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Solution:
Possible outcomes = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
No. of possible outcomes = 2 × 2 × 2 = 8
(i) Prob(all heads) = \(\frac{1}{8}\)
(ii) Atleast one tail = {(THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
Prob(atleast one tail) = \(\frac{7}{8}\)
(iii) Atmost one head = {(HTT), (THT), (TTH), (TTT)}
∴ Prob(atmost one head) = \(\frac{4}{8}=\frac{1}{2}\)
(iv) Atmost two tail = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT)}
∴ Prob(atmost two tail) = \(\frac{7}{8}\)

Question 9.
Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Answer:
1st dice A = {1, 2, 3, 4, 5, 6}
2nd dice B = {1, 1, 2, 2, 3, 3}
Sample Space (S) = {(1, 1), (1, 1), (1, 2), (1, 2), (1, 3), (1, 3), (2, 1), (2, 1), (2, 2), (2, 2), (2, 3), (2, 3), (3, 1), (3, 1), (3, 2), (3, 2), (3, 3), (3, 3), (4, 1), (4, 1), (4, 2), (4, 2), (4, 3), (4, 3), (5, 1), (5, 1), (5, 2), (5, 2), (5, 3), (5, 3),(6, 1), (6, 1), (6, 2), (6, 2), (6, 3), (6, 3)}
n(S) = 36
(i) Let A₁ be the event of getting sum is 2
A1 = {(1, 1) (1, 1)}
n(A₁) = 2
\(P\left(A_{1}\right)=\frac{n\left(A_{1}\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)
(ii) Let A₂ be the event of getting a sum is 3.
A₂ = {(1, 2) (1, 2) (2, 1) (2, 1)}
n(A₂) = 4
\(P\left(A_{2}\right)=\frac{4}{36}=\frac{1}{9}\)
(iii) Let A₃ be the event of getting a sum is 4.
A₃ = {(1, 3) (1, 3) (2, 2) (2, 2) (3, 1) (3, 1)}
n(A₃) = 6
\(P\left(A_{3}\right)=\frac{6}{36}=\frac{1}{6}\)
(iv) Let A₄ be the event of getting a sum is 5.
A₄ = {(2, 3) (2, 3) (3, 2) (3, 2) (4, 1) (4, 1)}
n(A₄) = 6
\(P\left(A_{4}\right)=\frac{6}{36}=\frac{1}{6}\)
(v) Let A₅ be the event of getting a sum is 6.
A₅ = {(3, 3) (3, 3) (4, 2) (4, 2) (5, 1) (5, 1)}
n(A₅) = 6
\(P\left(A_{5}\right)=\frac{6}{36}=\frac{1}{6}\)
(vi) Let A₆ be the event of getting a sum is 7.
A₆ = {(4, 3) (4, 3) (5, 2) (5, 2) (6, 1) (6, 1)}
n(A₆) = 6
\(P\left(A_{6}\right)=\frac{6}{36}=\frac{1}{6}\)
(vii) Let A₇ be the event of getting a sum is 8.
A₇ = {(5, 3) (5, 3) (6, 2) (6, 2)}
n(A₇) = 4
\(P\left(A_{7}\right)=\frac{4}{36}=\frac{1}{9}\)
(viii) Let A₈ be the event of getting a sum is 9.
A₈ = {(6, 3) (6, 3)}
n(A₈) = 2
\(P\left(A_{8}\right)=\frac{2}{36}=\frac{1}{18}\)

Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Answer:
Sample space (S) = 5 + 6 + 7 + 8
n(S) = 26
(i) Let A be the event of getting a white ball
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(P(A)=\frac{6}{26}=\frac{3}{13}\)
(ii) Let A be the event of getting a black ball
n(A) = 8
\(P(A)=\frac{n(A)}{n(S)}=\frac{8}{26}\)
Let B be the event of getting a red ball
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{26}\)
Probability of getting black or red ball
P(A ∪ B) = P (A) + P (B)
= \(\frac{8}{26}+\frac{5}{26}=\frac{13}{26}=\frac{1}{2}\)
(iii) Not white probability of getting white ball
P(A) = \(\frac{3}{13}\) from (i)
Probability of not getting white ball P(\(\bar{A}\)) = 1 – P(A)
\(1-\frac{3}{13}=\frac{13-3}{13}=\frac{10}{13}\)
(iv) Probability of getting a white ball.
P(A) = \(\frac{6}{26}\) (from 1)
Let B be the event of getting a black ball
n(B) = 8
\(P(B)=\frac{n(B)}{n(S)}=\frac{8}{26}\)
P(A ∪ B) = P(A) + P(B) = \(\frac{6}{26}+\frac{8}{26}=\frac{14}{26}\)
Probability of neither white nor black P(A ∪ B)’ = 1 – P(A ∪ B)
= \(1-\frac{14}{26}\)
= \(\frac{26-14}{26}=\frac{12}{26}=\frac{6}{13}\)

Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is \(\frac{3}{8}\) then, find the number of defective bulbs.
Answer:
Let the number of defective bulbs be “x”
Sample space (S) = 20 + x
n(S) = 20 + x
Let A be the event of getting to be defective
n(A) = x
\(P(A)=\frac{n(A)}{n(S)}\)
⇒ \(\frac{3}{8}=\frac{x}{20+x}\)
⇒ 8x = 3(20 + x) = (60 + 3x)
⇒ 8x – 3x = 60
⇒ 5x = 60
⇒ x = \(\frac{60}{5}\)
⇒ x = 12
Number of defective bulbs = 12

Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Answer:
King diamond + Queen diamonds = 1 + 1 = 2 …….(1)
Queen hearts + Jack of hearts = 1 + 1 = 2 …….(2)
Jack spade + King of spades =1 + 1 = 2 …….(3)
Remaining number of cards = 52 – (6)
n(S) = 46
(i) Let A be the event of getting a clavor
n (A) = 13
\(P(A)=\frac{n(A)}{n(S)}=\frac{13}{46}\)
(ii) Let B be the event of getting a queen of red card
n(B) = 2
But the above two cards are removed from (1) and (2)
n(B) = 0
\(P(B)=\frac{n(B)}{n(S)}=\frac{0}{46}=0\)
(iii) Let B be the event of getting a king of black card
n(B) = (2 – 1) [from (3) one black card is removed]
n (B) = 1
\(P(B)=\frac{n(B)}{n(S)}=\frac{1}{46}\)

Question 13.
Some boys are playing a game, in which the stone thrown by them landing in a circular region given in the figure is considered as win and landing other than the circular region is considered as a loss. What is the probability to win the game?
Answer:

Area of a rectangle = l × b sq. feet = 3 × 4 sq. feet = 12 sq. feet
sample space (S) = 12
n(S) = 12
Let A be the event of getting the stone landing in a circular region
n(A) = Area of a circle
= πr²
= π × 1 × 1 (radius of a circle = 1 feet)
= π

Probability to win the game = \(\frac{11}{42}\) (or) \(\frac{157}{600}\)

Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?
Answer:
Sample space (S) = 6 × 6 = 36
n(S) = 36
[priya and Amuthan are visiting a particular shop in any one of 6 days is 6 × 6 = 36]
(i) Let A be the event of getting both are shopping on the same day
A = {(Mon, Mon) (Tue, Tue) (Wed, Wed) (Thu, Thu) (Fri, Fri) (Sat, Sat)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of shopping in different days.
n(B) = 36 – 6 = 30
\(P(B)=\frac{n(B)}{n(S)}\)
\(=\frac{30}{36}=\frac{5}{6}\)
(iii) Let C be the event of shopping consecutive days
C = {(Mon, Tue) (Tue, Wed) (Wed, Thu) (Thu, Fri) (Fri, Sat)}
n(C) = 5
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}\) = \(\frac{5}{36}\)

Question 15.
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Answer:
Sample space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting double entry fee (only getting 3 heads)
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting her entry fee (one or two heads to show)
n(B) = Probability of one head + Probability of 2 head
= \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\)
(iii) To loss the entry means not getting the head (only tail)
n(C) = 1
\(P(C)=\frac{n(C)}{n(S)}=\frac{1}{8}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.2

Question 1.
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer:
Standard deviation of a data (σ) = 6.5
Mean of the data (\(\bar{x}\)) = 12.5
Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100 \%\)
= \(\frac{6.5}{12.5} \times 100 \%=52 \%\)
Coefficient of variation = 52%

Question 2.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Answer:
Standard deviation (σ) = 1.2
Coefficient of variation = 25.6

Question 3.
If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Answer:
Mean (\(\bar{x}\)) = 15
Co efficient of variation = 48

Question 4.
If n = 5, \(\bar{x}\) = 6, Σx² = 765, then calculate the coefficient of variation.
Answer:

Question 5.
Find the coefficient of variation of 24, 26, 33, 37, 29, 31.
Answer:
Arrange in ascending order we get 24, 26, 29, 31, 33, 37
Assumed mean = 29

Question 6.
The time taken (in minutes) to complete homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Answer:
Arrange in ascending order we get, 38, 40, 43, 44, 46, 47, 49, 53.
Assumed mean = 46


= \(\frac{453}{45}\)%
= 10.066
Coefficient of variation = 10.07%

Question 7.
The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation of 4.6 and 2.4 respectively. Who is more consistent in performance?
Answer:
Total marks scored by sathya = 460
Total marks scored by vidhya = 480
Number of subjects = 5
Mean marks of sathya = \(\frac{460}{5}\)
\(\bar{x}\) = 92%
Given standard deviation, (σ) = 4.6

Vidhya coefficient of variation is less than Sathya.
Vidhya is more consistent.

Question 8.
The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.

Which of the three subjects shows the highest variation and which shows the lowest variation in marks?
Answer:
(i) Mathematics:
Mean (\(\bar{x}\)) = 56
Standard deviation (σ) = 12
Coefficient variation (CV₁) = \(\frac{\sigma}{\bar{x}} \times 100=\frac{12}{56} \times 100\)

Science shows the highest variation
Social science shows the lowest variation

Question 9.
The temperature of two cities A and B in the winter season are given below.

Find which city is more consistent in temperature changes?
Answer:
(i) city A:
Assumed mean = 22




C.V of city A < C.V of city B
City A is more consistent in temperature change.

Students can download Maths Chapter 7 Mensuration Unit Exercise 7 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Unit Exercise 7

Question 1.
The barrel of a fountain-pen cylindrical in shape is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one-fifth of a litre?
Answer:

Question 2.
A hemispherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litres per second. How much time will it take to empty the tank completely?
Answer:
Radius of the hemispherical tank = 1.75 m
Volume of the tank = \(\frac{2}{3} \pi r^{3}\) cu.units

Time taken = \(\frac{11229.17}{7}\) = 1604.17 seconds = 26.74 minutes = 27 minutes (approximately)

Question 3.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Answer:
Radius of a cone = Radius of a hemisphere = r unit

Height of a cone = r units
(height of the cone = radius of a hemisphere)
Maximum volume of the cone

Question 4.
An oil funnel of the tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion by 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Answer:

Total height of oil funnel = 22 cm
Height of the cylindrical portion = 10 cm
Height of the frustum (h) = 22 – 10 = 12 cm
Radius of the cylindrical portion = 4 cm
Radius of the bottom of the frustum = 4 cm
Top radius of the funnel (frustum) = \(\frac{18}{2}\) = 9 cm
Area of the tin sheet required = C.S.A of the frustum + C.S.A of the cylinder
= π (R + r) l + 2πrh sq. units.
= [π(9 + 4) \(\sqrt{12^{2}+(9-4)^{2}}\) + 2π × 4 × 10] cm²
= π [13 × \(\sqrt{144+25}\) + 25 + 80] cm²
= \(\frac{22}{7}\) [13 × 13 + 80] cm²
= \(\frac{22}{7}\) [169 + 80] cm²
= \(\frac{22}{7}\) × 249 cm²
= 782.57 cm²
Area of sheet required to make the funnel = 782.57 cm²

Question 5.
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer:
Radius of the cylinder = \(\frac{4.5}{2}\) cm
Height of the cylinder = 10 cm
Volume of the cylinder = πr²h cu. units


Number of coins = 450

Question 6.
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and the whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of a solid cylinder.
Answer:
External radius of the hollow cylinder R = 4.3 cm
Internal radius of the hollow cylinder r = 1.1 cm
Length of the cylinder (h) = 4 cm
Length of the solid cylinder (H) = 12 cm
Let the radius of the solid cylinder be “x”
Volume of the solid cylinder = Volume of the hollow cylinder

Diameter of the solid cylinder = 2 × 2.4 = 4.8 cm

Question 7.
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m.
Answer:
Slant height of a frustum (l) = 4 m
Perimeter of the top part = 18 m
2πR = 18


Cost of painting = ₹ 100 × 68 = ₹ 6800

Question 8.
A hemispherical hollow bowl has material of volume cubic \(\frac{436 \pi}{3}\) cubic cm. Its external diameter is 14 cm. Find its thickness.
Answer:
External radius of a hemisphere (R) = 7 cm
Volume of a hemi-spherical bowl = \(\frac{436 \pi}{3}\) cm³

Internal radius = 5 cm
Thickness of the hemisphere = (7 – 5) cm = 2 cm

Question 9.
The volume of a cone is 1005\(\frac{5}{7}\) cu. cm. The area of its base is 201\(\frac{1}{7}\) sq. cm. Find the slant height of the cone.
Answer:
Area of the base of a cone = 201\(\frac{1}{7}\) sq. cm

Question 10.
A metallic sheet in the form of a sector of a circle of radius 21 cm has a central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Answer:

Radius of a cone (r) = 21 cm
Central angle (θ) = 216°
Let “R” be the radius of a cone
Circumference of the base of a cone = arc length of the sector
2πR = \(\frac{\theta}{360} \times 2 \pi r\)

Students can download Maths Chapter 7 Mensuration Ex 7.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.5

Multiple Choice Questions

Question 1.
The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is
(1) 60π cm²
(2) 68π cm²
(3) 120π cm²
(4) 136π cm²
Solution:
(4) 13671 cm²
Hint:
Here, h = 15 cm, r = 8 cm

C.S.A of a cone = πrl sq. units. = π × 8 × 17 = 136π cm³

Question 2.
If two solid hemispheres of same base radius r units are joined together along with their bases, then the curved surface area of this new solid is
(1) 4πr² sq. units
(2) 6πr² sq. units
(3) 3πr² sq. units
(4) 8πr² sq. units
Answer:
(1) 4πr² sq. units
Hint:
When you joined two hemispheres together, the solid sphere is formed
C.S.A of the new solid = C.S.A of a sphere = 4πr² sq. units.

Question 3.
The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be
(1) 12 cm
(2) 10 cm
(3) 13 cm
(4) 5 cm
Solution:
(1) 12 cm
Hint:
Here r = 5 cm and l = 13 cm

Question 4.
If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is _________
(1) 1 : 2
(2) 1 : 4
(3) 1 : 6
(4) 1 : 8
Answer:
(2) 1 : 4
Hint:
Let the radius of the cylinder be “r” and the height be “h”
Radius of the new cylinder = \(\frac{r}{2}\) (Height will be same)
Volume of the new cylinder : Volume of the original cylinder
= \(\pi r_{1}^{2} h: \pi r_{2}^{2} h\) (πh is same)
= \(r_{1}^{2}: r_{2}^{2}\)
= \(\left(\frac{r}{2}\right)^{2}: r^{2}\)
= \(\frac{r^{2}}{4}: r^{2}=\frac{1}{4}: 1\)
= 1 : 4

Question 5.
The total surface area of a cylinder whose radius is \(\frac{1}{3}\) of its height is _______
(1) \(\frac{9 \pi h^{2}}{8}\) sq. units
(2) 24πh² sq.units
(3) \(\frac{8 \pi h^{2}}{8}\) sq.units
(4) \(\frac{56 \pi h^{2}}{8}\) sq.units
Answer:
(3) \(\frac{8 \pi h^{2}}{8}\) sq.units
Hint:
Let the height of the cylinder be “h”
Radius of the cylinder = \(\frac{1}{3}\) h
T.S.A of the cylinder = 2πr(h + r)

Question 6.
In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is
(1) 5600π cm³
(2) 11200π cm³
(3) 56π cm³
(4) 3600π cm³
Solution:
(2) 112007π cm³
Hint:
Here, let the external radius be “R” and the internal radius be “r”
R + r = 14 ……(1)
Width (R – r) = 4 ……(2)
Height of the hollow cylinder = 20 cm
Volume of the hollow cylinder = πh × (R² – r²)
= πh(R + r) (R – r)
= π × 20 (14) × 4
= π × 1120
= 1120π cm³

Question 7.
If the radius of the base of a cone is tripled and the height is doubled then the volume is ______
(1) made 6 times
(2) made 18 times
(3) made 12 times
(4) unchanged
Answer:
(2) made 18 times
Hint:
Radius of a cone = r
Height of a cone = h
Volume of the cone = \(\frac{1}{3}\) πr²h cu. units
When the radius is increased three-time (tripled) and the height is doubled
Radius is 3r and the height is 2h
Volume of the new cone

Volume is increased 18 times.

Question 8.
The total surface area of a hemisphere is how many times the square of its radius.
(1) π
(2) 4π
(3) 3π
(4) 2π
Solution:
(3) 3π
Hint:
T.S.A of the hemisphere = 3πr²
The square of the radius is 3π times.

Question 9.
A solid sphere of radius x cm is melted and cast into a shape of a solid cone of the same radius. The height of the cone is _______
(1) 3x cm
(2) x cm
(3) 4x cm
(4) 2x cm
Answer:
(3) 4x cm
Hint:
Radius of a sphere = Radius of a cone = x cm
Volume of a cone = Volume of a sphere

Question 10.
A frustum of a right circular cone is of height 16cm with radii of its ends as 8cm and 20cm. Then, the volume of the frustum is
(1) 3328π cm³
(2) 3228π cm³
(3) 3240πcm³
(4) 3340π cm³
Solution:
(1) 3328π cm³
Hint:
Here, h = 16 cm, r = 8 cm, R = 20 cm
Volume of the frustum

Question 11.
A shuttlecock used for playing badminton has the shape of the combination of ______
(1) a cylinder and a sphere
(2) a hemisphere and a cone
(3) a sphere and a cone
(4) frustum of a cone and a hemisphere
Answer:
(4) frustum of a cone and a hemisphere
Hint:

Question 12.
A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is _______
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer:
(1) 2 : 1
Hint:
Volume of the first sphere : Volume of second sphere = 8 : 1

Question 13.
The volume (in cm³) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is ________
(1) \(\frac{4}{3} \pi\)
(2) \(\frac{10}{3} \pi\)
(3) 5π
(4) \(\frac{20}{3} \pi\)
Answer:
(1) \(\frac{4}{3} \pi\)
Hint:
Radius of the sphere = 1 cm
Volume of the Sphere = \(\frac{4}{3}\) πr³ cu. units
= \(\frac{4}{3}\) × π × 1 × 1 × 1 cm³
= \(\frac{4}{3}\) π cm³

Question 14.
The height and radius of the cone of which the frustum is a part are h₁ units and r₁ units respectively. Height of the frustum is h₂ units and the radius of the smaller base is r₂ units. If h₂ : h₁ = 1 : 2 then r₂ : r₁ is ______
(1) 1 : 3
(2) 1 : 2
(3) 2 : 1
(4) 3 : 1
Answer:
(2) 1 : 2
Hint:
h₂ : h₁ = 1 : 2
h₁ : h₂ = 2 : 1
Ratio of their volumes

Volume is 2 : 1 the ratio of their radius also 2 : 1
r₁ : r₂ = 2 : 1 But r₂ : r₁ = 1 : 2

Question 15.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is
(1) 1 : 2 : 3
(2) 2 : 1 : 3
(3) 1 : 3 : 2
(4) 3 : 1 : 2
Solution:
(4) 3 : 1 : 2
Hint:
Volume of (cylinder : cone : sphere)

Students can download Maths Chapter 7 Mensuration Ex 7.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.4

Question 1.
An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder.
Answer:
Sphere – Radius r₁ = 12 cm
Cylinder – Radius r₂ = 8 cm
h₂ = ?
Volume of cylinder = Volume of sphere melted

∴ Height of the cylinder made = 36 cm.

Question 2.
Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.
Answer:
Length of the rectangular tank (l) = 50 m = 5000 cm
Width of the rectangular tank (b) = 44 m = 4400 cm
Level of water in the tank (h) = 21 cm
Volume of the tank = l × b × h cu. units = 5000 × 4400 × 21 cm³
Radius of the pipe (r) = 7 cm
Speed of the water = 15 km/hr.
(h) = 15000 × 100 cm / hr.
Volume of water flowing in one hour

Question 3.
A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius x r units. Find the height of water in the cylindrical flask.
Answer:
Radius of the conical flask = r units
Height of the conical flask = h units
Volume of the conical flask = \(\frac{1}{3} \pi r^{2} h\) cu.units
Radius of the cylindrical flask = x r units
Let the height of the cylindrical flask be “H” units
Volume of the cylindrical flask = Volume of the Conical flask

Height of the cylindrical flask = \(\frac{h}{3 x^{2}}\) units

Question 4.
A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter.
Answer:
Radius of a cone (V) = 7 cm
Height of a cone (h) = 8 cm
External radius of the hollow sphere (R) = 5 cm
Let the internal radius be “x”
Volume of the hollow sphere = Volume of the Cone


Internal diameter of the Hollowsphere = 2 × 3 = 6 cm.

Question 5.
Seenu’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2 m × 1.5 m × 1 m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.
Answer:
Length of the cuboid tank (l) = 2 cm = 200 cm
Breadth of the cuboid tank (b) = 1.5 cm = 150 cm
Height of the tank (h) = 1 m = 100 cm
Volume of the cuboid = l × b × h cu. units
= 200 × 150 × 100 cm³
= 30,00,000 cm³
Radius of the tank (r) = 60 cm
Height of the tank (h) = 105 cm
Volume of the cylindrical tank = πr²h cu. units
= \(\frac{22}{7}\) × 60 × 60 × 105 cm³
= 22 × 60 × 60 × 15 cm³
= 1188000 cm³
Volume of water left in the sump = Volume of the sump – Volume of the tank
= 3000000 – 1188000 cm³
= 1812000 cm³

Question 6.
The internal and external diameter of a hollow hemispherical shell is 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of the cylinder.
Answer:
Internal radius of the shell (r) = 3 cm
External radius of the shell (R) = 5 cm
Radius of the cylinder (r) = 7 cm
Let the height of the cylinder be “h”
Volume of the cylinder = Volume of the hemispherical shell

Height of the cylinder = 1.33 cm

Question 7.
A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder.
Answer:
Radius of a sphere (r) = 6 cm
External radius of the cylinder (R) = 5 cm
Height of the cylinder (h) = 32 cm
Let the internal radius of the cylinder be ‘x’
Volume of the hollow cylinder = Volume of a sphere
πh (R² – r²) = \(\frac{4}{3}\) πr³
π × 32 (5 + x) (5 – x) = \(\frac{4}{3}\) × π × 6 × 6 × 6
32 (25 – x²) = 4 × 2 × 6 × 6
25 – x² = 9
x² = 25 – 9 = 16
x = √16 = 4
Thickness of the cylinder = 5 – 4 = 1 cm.

Question 8.
A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.
Answer:
Let the height of the cylinder be “h”
radius is 50% more than the height


From (1) and (2) we get,
Volume of the cylinder = Volume of the hemisphere
It is possible to transfer the full quantity from the bowl into the cylindrical vessel.
100 % of the juice can be transferred.

Students can download Maths Chapter 7 Mensuration Ex 7.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.3

Question 1.
A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 cm and the height of the vessel is 13 cm. Find the capacity of the vessel.
Answer:
Radius of a hemisphere = Radius of the cylinder

Question 2.
Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of the model that Nathan made.
Answer:
Radius of the cone = Radius of the cylinder

r = \(\frac{3}{2}\) cm
Height of the cone (H) = 2 cm
Height of the cylinder (h) = 12 – (2 + 2) cm = 8 cm
Volume of the model = Volume of the cylinder + Volume of 2 cones

Question 3.
From a solid cylinder whose height is 2.4 cm and the diameter 1.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm³.
Answer:
Radius of a cylinder = Radius of a cone r = 0.7 cm
Height of a cylinder = Height of a cone (h) = 2.4 cm

Volume of the remaining solid = Volume of the cylinder – Volume of a cone

Volume of the remaining soild = 2.46 cm³

Question 4.
A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.

Answer:
Radius of a cone = Radius of a hemisphere = Radius of a cylinder
r = 6 cm
Height of a cone (h) = 12 cm
Volume of the water displaced = Volume of the solid inside = Volume of the cone + Volume of the hemisphere

Volume of water displaced = 905. 14 cm³.

Question 5.
A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?
Answer:
Radius of a hemisphere = Radius of a Cylinder

r = \(\frac{3}{2}\) mm = 1.5 mm
Height of the cylinderical portion = 12 mm – (1.5 mm + 1.5 mm) = (12 – 3) mm = 9 mm
Volume of the capsule

Volume of the capsule = 77.8 cu. mm

Question 6.
As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.

Answer:
Side of a cube (a) = 7 cm
Radius of a hemisphere (r) = \(\frac{7}{2}\) cm
Surface area of the solid = T.S.A of the cube + C.S.A of the hemisphere – Area of the base of the hemisphere

Question 7.
A right circular cylinder just encloses a sphere of radius r units. Calculate
(i) the surface area of the sphere
(ii) the curved surface area of the cylinder
(iii) the ratio of the areas obtained in (i) and (ii).
Answer:
(i) Surface area of sphere = 4πr² sq. units

Question 8.
A shuttlecock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttlecock is 7 cm. Find its external surface area.
Answer:

Students can download Maths Chapter 7 Mensuration Ex 7.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.2

Question 1.
A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.

Answer:
Radius of the well (r₁) = 5 m
Depth of the well (h) = 14 m
Width of the embankment = 5 m
Outer radius (R) = 5 + 5 = 10 m
Let the height of the embankment be “H”
Volume of Earth in the embankment = Volume of the well
πH(R² – r²) = \(\pi r_{1}^{2} h\)
H(10² – 5²) = 5 × 5 × 14
H (100 – 25) = 5 × 5 × 14
H = \(\frac{5 \times 5 \times 14}{75}\) = 4.67 m
Height of the embankment = 4.67 m

Question 2.
A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the rise of the water in the glass?
Answer:
Radius of the cylindrical glass (r) = 10 cm
Height of the water (h) = 9 cm
Radius of the cylindrical metal (R) = 5 cm
Height of the metal (H) = 4 cm
Let the height of the water raised be “h”
Volume of the water raised in the cylinder = Volume of the cylindrical metal
πr²h = πr²H
10 × 10 × h = 5 × 5 × 4
h = \(\frac{5 \times 5 \times 4}{10 \times 10}\) = 1 cm
Raise of water in the glass = 1 cm

Question 3.
If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.
Answer:
Circumference of the wooden piece = 484 cm
2πr = 484
2 × \(\frac{22}{7}\) × r = 484 cm
r = \(\frac{484 \times 7}{2 \times 22}\)
r = 77 cm
Height of the wooden piece (h) = 105 cm
Volume of the conical wooden piece = \(\frac{1}{3} \pi r^{2} h\) cu.units
= \(\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 \mathrm{cm}^{3}\)
= 22 × 11 × 77 × 35 cm³
= 652190 cm³
Volume of the wooden piece = 652190 cm³

Question 4.
A conical container is fully filled with petrol. The radius is 10m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.
Answer:
The radius of the conical container (r) = 10 m
Height of the container (h) = 15 m

Question 5.
A right-angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.

Answer:
Three sides of a triangle are 6 cm, 8 cm and 10 cm.
Case (i): If the triangle is revolved about the side 6 cm, the cone will be formed with radius 6 cm and height 8 cm.
Volume of the cone = \(\frac{1}{3} \pi r^{2} h\) cu. units
= \(\frac{1}{3}\) × π × 6 × 6 × 8 = 96π cm³
Case (ii): If the triangle is revolved about the side 8 cm, the cone will be formed with radius 8 cm and height 6 cm.
Volume of the cone = \(\frac{1}{3}\) × π × 8 × 8 × 6 = 128π cm³
Difference in volume of the two solids = (128π – 96π) cm³ = 32π cm³ = 32 × \(\frac{22}{7}\) cm³ = 100.57 cm³
The difference in the volume of the two solids = 100.57 cm³

Question 6.
The volumes of two cones of same base radius are 3600 cm³ and 5040 cm³. Find the ratio of heights.
Answer:
Let the radius of the two cones be ‘r’
Let the height of the two cones be h₁ and h₂
Ratio of their volumes = 3600 : 5040 (÷ 10)
\(\frac{1}{3} \pi r^{2} h_{1}: \frac{1}{3} \pi r^{2} h_{2}\) = 360 : 504 (÷4)
h₁ : h₂ = 90 : 126 (÷3)
= 30 : 42 (÷3)
= 10 : 14 (÷2)
h₁ : h₂ = 5 : 7
Ratio of heights = 5 : 7

Question 7.
If the ratio of radii of two spheres is 4 : 7, find the ratio of their volumes.
Answer:
Let the ratio of their radii is r₁ : r₂
r₁ : r₂ = 4 : 7
Ratio of their volumes
V₁ : V₂ = \(\frac{4}{3} \pi r_{1}^{3}: \frac{4}{3} \pi r_{2}^{3}\)
= \(r_{1}^{3}: r_{2}^{3}\)
= 43 : 73
Ratio of their volumes = 64 : 343

Question 8.
A solid sphere and a solid hemisphere have an equal total surface area. Prove that the ratio of their volume is 3√3 : 4.
Answer:
Total surface area of a sphere = \(4 \pi r_{1}^{2}\) sq. units
Total surface area of a hemisphere = \(3 \pi r_{2}^{2}\) sq. units
Ratio of Total surface area = \(4 \pi r_{1}^{2}: 3 \pi r_{2}^{2}\)
1 = \(\frac{4 \pi r_{1}^{2}}{3 \pi r_{2}^{2}}\) (Same Surface Area)
1 = \(\frac{4 r_{1}^{2}}{3 r_{2}^{2}}\)

Ratio of their volumes = 3√3 : 4
Hence it is proved.

Question 9.
The outer and the inner surface areas of a spherical copper shell are 576π cm² and 324π cm² respectively. Find the volume of the material required to make the shell.
Answer:
Outer surface area of a spherical shell = 576π cm²
4πR² = 576π
4 × R² = 576
R² = \(\frac{576}{4}\) = 144
R = √144 = 12 cm
Inner surface area of a spherical shell = 324π cm²
4πr² = 324π
4r² = 324
r² = 81
r = √81 = 9
Volume of the material required = Volume of the hollow hemisphere = \(\frac{4}{3}\) π(R³ – r³) cm³

Volume of the material required = 4186.29 cm³

Question 10.
A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of ₹ 40 per litre.
Answer:
Height of the frustrum (h) = 16 cm
Radius of the upper part (R) = 20 cm
Radius of the lower part (r) = 8 cm
Volume of the frustum

Cost of milk in the container = 10.459 × 40 = ₹ 418.36
Cost of the milk = ₹ 418.36

Students can download Maths Chapter 7 Mensuration Ex 7.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.1

Question 1.
The radius and height of a cylinder are in the ratio 5 : 7 and its curved surface area is 5500 sq.cm. Find its radius and height.
Answer:
Let the radius be 5x and the height be 7x
C.S.A of a cylinder = 5500 sq.cm.
2πrh = 5500
2 × \(\frac{22}{7}\) × 5x × 7x = 5500
2 × 22 × 5 × x² = 5500
x² = \(\frac{5500}{2 \times 22 \times 5}\)
x² = 25 cm
x = 5 cm
Radius of the cylinder = 5 × 5 = 25 cm
Height of the cylinder = 7 × 5 = 35 cm

Question 2.
A solid iron cylinder has total surface area of 1848 sq.m. Its curved surface area is five-sixth of its total surface area. Find the radius and height of the iron cylinder.
Answer:
T.S.A of the cylinder =1848 sq.cm
2πr(h + r) = 1848 ……. (1)
Curved surface area = \(\frac{5}{6}\) × 1848 sq.cm
2πrh = 5 × 308
2πrh = 1540 sq.m ……… (2)
Substitute the value of2πrh in (1)
2πr(h + r) = 1848
2πrh + 2πr² = 1848
1540 + 2πr² = 1848
2πr² = 1848 – 1540
2 × \(\frac{22}{7}\) × r² = 308
r² = \(\frac{308 \times 7}{2 \times 22}\) = 49
r = 7
Radius of the cylinder = 7m
2πrh = 1540
2 × \(\frac{22}{7}\) × 7 × h = 1540
h = \(\frac{1540}{2 \times 22}\) = 35 m
Radius of the cylinder = 7 m
Height of the cylinder = 35 m

Question 3.
The external radius and the length of a hollow wooden log are 16 cm and 13 cm respectively. If its thickness is 4 cm then find its T.S.A.
Answer:
External radius of the wooden log (R) = 16 cm
Thickness = 4 cm
Internal radius (r) = 16 – 4 = 12 cm
Length of the wooden log (h) = 13 cm
T.S.A of the hollow cylinder = 2π (R + r) (R – r + h) sq.cm
= 2 × \(\frac{22}{7}\) × (16 + 12) (16 – 12 + 13) sq.cm
= 2 × \(\frac{22}{7}\) × 28 × 17 sq.cm
= 2 × 22 × 4 × 17 sq.cm.
= 2992 sq.cm.
T.S.A of the hollow wooden = 2992 sq.cm.

Question 4.
A right angled triangle PQR where ∠Q = 90° is rotated about QR and PQ. If QR = 16 cm and PR = 20 cm, compare the curved surface areas of the right circular cones so formed by the triangle.
Answer:

In the Right Triangle
QP² = PR² – QR²= 20² – 16² = 400 – 256 = 144
QP = √144 = 12 cm
When PQ is rotated r = 12, l = 20
C.S.A of the cone = πrl sq. units = π × 12 × 20 cm² = 240π cm²
When QR is rotated r = 16, l = 20
C.S.A of the cone = nrl sq. units = π × 16 × 20 = 320π cm²
C.S.A. of a cone when rotated about QR is larger.

Question 5.
4 persons live in a conical tent whose slant height is 19 cm. If each person requires 22 cm² of the floor area, then find the height of the tent.
Answer:
Slant height of a cone (r) = 19 cm
Floor area for 4 persons = 4 × 22 cm²
πr² = 88 cm

Height of the tent = 18.25 cm

Question 6.
A girl wishes to prepare birthday caps in the form of right circular cones for her birthday party, using a sheet of paper whose area is 5720 cm², how many caps can be made with radius 5 cm and height 12 cm.
Answer:
Radius of a cap (r) = 5 cm

Question 7.
The ratio of the radii of two right circular cones of the same height is 1 : 3. Find the ratio of their curved surface area when the height of each cone is 3 times the radius of the smaller cone.
Answer:
Let the radius of the first cone be ‘x’ and the Height of the cone be 3x

Question 8.
The radius of a sphere increases by 25%. Find the percentage increase in its surface area.
Answer:
Let the radius of the be “r”
Surface area of the sphere = 4πr² sq.units …….. (1)
If the radius is increased by 25%



Percentage of increase in surface area = 56.25 %

Question 9.
The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost to paint the vessel all over at ₹ 0.14 per cm².
Answer:

Question 10.
The frustum shaped outer portion of the table lamp has to be painted including the top part. Find the total cost of painting the lamp if the cost of painting 1 sq.cm is ₹ 2.

Answer:
The slant height of the frustum (l)


Cost of painting = ₹ 678.86 × 2 = ₹ 1357.72

Students can download 10th Social Science Economics Chapter 5 Industrial Clusters in Tamil Nadu Questions and Answers, Notes, Samacheer Kalvi 10th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Social Science Solutions Economics Chapter 5 Industrial Clusters in Tamil Nadu

Samacheer Kalvi 10th Social Science Industrial Clusters in Tamil Nadu Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
‘The Detroit of Asia’ is
(a) Tuticorin
(b) Coimbatore
(c) Chennai
(d) Madurai
Answer:
(c) Chennai

Question 2.
Pumpsets and motors are produced mostly in …………….
(a) Salem
(b) Coimbatore
(c) Chennai
(d) Dharampuri
Answer:
(b) Coimbatore

Question 3.
Tuticorin is known as:
(a) Gateway of India
(b) Gateway of Tamil Nadu
(c) Pump city
(d) None of these
Answer:
(b) Gateway of Tamil Nadu

Question 4.
……………… are an essential aspect of a nation’s development.
(a) Agriculture
(b) Industry
(c) Railway
(d) none of these
Answer:
(a) Agriculture

Question 5.
Tiruppur is known for:
(a) Leather tanning
(b) Lock making
(c) Knitwear
(d) Aigo-processing
Answer:
(c) Knitwear

Question 6.
Along with Ambur and Vaniyambadi ………….. is also a centre for leather goods exports.
(a) Chennai
(b) Sivakasi
(c) Coimbatore
(d) Madurai
Answer:
(a) Chennai

Question 7.
IT means
(a) Indian Technology
(b) Information Technology
(c) Institute of Technology
(d) Initiative Technology
Answer:
(b) Information Technology

Question 8.
A successful industrial cluster entirely created by the Tamil Nadu is ………………
(a) Hosur
(b) Dindigul
(c) Kovilpatti
(d) Thirunelveli
Answer:
(a) Hosur

Question 9.
SIPCOT was formed in the year:
(a) 1972
(b) 1976
(c) 1971
(d) 1978
Answer:
(c) 1971

Question 10.
Which is the Tamil Nadu Industrial Development Agency?
(a) SIPCOT
(b) TANSIDCO
(c) TIDCO
(d) all of these
Answer:
(d) all of these

II. Fill in the blanks

1. ………….. are very important in the modern economic activates of man.
2. ………….. are groups of firms in a defined geographic area that share common market and technologies.
3. Hundred of leather and tannery facilities are located around ………….. District inTamilNadu.
4. ………….. is fondly calls as ‘Little Japan’.
5. Special Economic Zones policy was introduced on in …………..
6. ………….. is an innovator of new ideas and business processes.

Answers:

1. Industrialisation
2. Industrial Clusters
3. Vellore
4. Sivakasi
5. April 2000
6. Entrepreneur

III. Choose the correct statement

Question 1.
(i) Entrepreneurship promotes capital formation by mobilising the idle saving of the public.
(ii) They are not provide large scale employment to artisan, technically qualified persons and professionals.
(iii) Entrepreneurs help the country to increase the GDP and per capitals income.
(iv) Entrepreneurs not promote country’s export trade.
(a) (i) and (iv) is correct
(b) (i) only correct
(c) (iii) and (iv) is correct
(d) (i) and (iii) is correct
Answer:
(d) (i) and (iii) is correct

IV. Pick out odd one

Question 1.
Which one of the following is not having leather factories?
(a) Ranipet
(b) Bharmapuri
(c) Ambur
(d) Vaniyambadi
Answer:
(b) Bharmapuri

Question 2.
Which one of the following is not a industrial developing agency?
(a) TIDCO
(b) SIDCO
(c) MEPG
(d) SIPCOT
Answer:
(c) MEPG

V. Match the following

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

VI. Write Short Answer

Question 1.
Why should a developing economy diversify out of agriculture?
Answer:
As an economy grows and incomes increase, consumers tend to spend a lesser share of their income on products from the agricultural sector. There are limits to the ability of agriculture to absorb labour due to the declining marginal productivity of land. Due to this, there is a need for an economy’s production and employment base to diversity away from agriculture.

Question 2.
Why are wages low in the agricultural sector?
Answer:
Labour productivity cannot increase in the agricultural sector as the marginal productivity of land goes on decreasing. Therefore wages remain low resulting in poverty.

Question 3.
What is meant by an industrial cluster?
Answer:
Industrial clusters are groups of firms in a defined geographic area that share common markets, technologies and skill requirements. An important aspect of clusters is the nature of inter-firm networks and interactions.

Question 4.
What are the routes for cluster formation?
Answer:

1. When artisans settle in one region and slowly spread their art in other regions.
2. When a large firm is established, to take care of its inputs and service requirements a cluster of firms may emerge.
3. Governments initiative to encourage Industrial sector using raw materials from a region.

Question 5.
Mention the 3 areas of policy-making that helped Tamil Nadu become one of the most industrialised states in the country.
Answer:
(a) Tamilnadu Industrial Policy – 2014
(b) Tamilnadu SEZs Policy
(c) Tamilnadu Biotechnology Policy – 2014

Question 6.
Mention any three-industrial development agencies in Tamil Nadu and their role.
Answer:
SIPCOT – State Industries Promotion Corporation of Tamil Nadu.

To promote Industrial growth by setting up Industrial Estates.

TIIC – Tamil Nadu Industrial Investment Cooperation Limited

To provide low – cost financial support for both setting up new firms and also to expand.

TIDCO – Tamil Nadu Industrial Development Corporation.

To promote Industrial growth by setting up Industrial Estate

Question 7.
What are the problems of industrialization currently in Tamil Nadu?
Answer:
To begin with, some clusters, especially chemicals, textiles and leather clusters, tend to generate a lot of polluting effluents that affect health. The effluents also pollute water bodies ‘ into which effluents are let into and also adjoining agricultural lands.

Second, employment generation potential has declined because of use of frontier technologies because of the need to compete globally. Quality of employment also has suffered in recent years as most workers are employed only temporarily. This issue too requires urgent attention among policymakers.

Question 8.
What is Meant by Entrepreneur?
Answer:
Entrepreneur is the person one who possess management skills, strong team building abilities and essential leadership qualities to manage a business.

Question 9.
What is Entrepreneurship?
Answer:
Entrepreneurship is a process of an action of an entrepreneur who undertakes to establish his entreprise. It is the ability to create and build something.

VII. Write brief Answer

Question 1.
What are the contributions of industrialization to development?
Answer:

1. Industries contribute by producing inputs such as fertilizers and tractors to agriculture thereby help them to increase productivity.
2. Services like Banking, transport and trade are dependent on the production of Industrial goods.
3. By using modem methods of production Industries contribute to better productivity and hence lower cost of production:
4. Industries helps to absorb the labour force coming out of agriculture.
5. By using modem technology, labour productivity increases, which help workers to get higher wages.
6. Increased income of the people lead to more demand for goods and services.
7. By producing more Industrial products, exports increases, thereby generate more foreign exchange.

Question 2.
Write a note on history of industrialisation in Tamil Nadu.
Answer:
Industrialisation in the Colonial Period

• The introduction of cotton cultivation in western and southern Tamil Nadu by the colonial government led to the emergence of a large-scale textile sector in these parts, which involved ginning, pressing, spinning and weaving operations.
• Introduction of railways also expanded the market for cotton yam and helped develop the sector.
• There was increase in trade during this period which led to industrial development. The two active ports in the region were Chennai and Tuticorin.
• In Western Tamil Nadu, the emergence of textiles industries also led to demand and starting
  of textile machinery industry in the region.

Post-Independence to early 1990s:

• After independence, several large enterprises were set up by both the central and state governments.
• The Integral Coach Factory in Chennai made railway coaches and the Bharat Heavy Electricals Limited (BHEL) in Tiruchirapalli manufactured boilers and turbines.
• Ashok Motors and Standard Motors together helped form an automobile cluster in the Chennai region.
• The 1970s and 1980s saw the setting up of emergence of power loom weaving clusters in the Coimbatore region as well as expansion of cotton knitwear cluster in Timppur and home furnishings cluster in Kamr.
• The Hosur industrial cluster is a successful case of how such policy efforts to promote industrial estates helped develop industries in a backward region.

Industrialization in Tamil Nadu – Liberalization Phase:

• The final phase of industrialisation is the post-reforms period since the early 1990s.
• Because of trade liberalisation measures, exports of textiles, home furnishings and leather products began to grow rapidly.
• Efforts to attract investments led to entry of leading multinational firms (MNCs) into the state, especially in the automobile sector.
• Chennai region also emerged as a hub for electronics industry with MNCs such as Nokia, Foxconn, Samsung and Flextronics opening plants on the city’s outskirts.
• A significant share of these investments has come up in special economic zones in the districts bordering Chennai.
• The major industries are automobiles, autocomponents, light and heavy engineering, machinery, cotton, etc.
• This diffused process of industrialisation and corresponding urbanisation has paved the way for better rural-urban linkages in Tamil Nadu than in most other states.

Question 3.
What are the important characteristics of successful industrial clusters? .
Answer:
The following are the chief characteristics of a successful cluster.

1. Geographical proximity of Small and Medium Enterprises (SMEs)
2. Sectoral specialisation
3. Close inter firm collaboration
4. Inter – firm competition based on innovation
5. A Socio – Cultural Identity which facilitates trust.
6. Multi – skilled workforce
7. Active self – help organisations and
8. Supportive regional and municipal governments.
9. Through Competition, they are forced to become more efficient.

Question 4.
Write about the Textile industry cluster in Tamil Nadu?
Answer:
Textile Clusters:
Tamil Nadu is home to the largest textiles sector in the country. Because of the development of cotton textile industry since the colonial period, Coimbatore often referred as the “Manchester of South India”. At present, most of the spinning mills have moved to the smaller towns and villages at a radius over 100 to 150 km around the Coimbatore city. Tamil Nadu is the biggest producer of cotton yam in the country.

Palladam and Somanur, small towns near Coimbatore and the villages near these towns, are home to a dynamic powerloom weaving cluster as well. Powerloom is however more widespread with Erode and Salem region too having a large number of power loom units.

Tiruppur is famous for clustering of a large number of firms producing cotton knitwear. It accounts for nearly 80% of the country’s cotton knitwear exports and generates employment in the range of over three lakh people since the late 1980s. It is also a major producer for the domestic market. Because of its success in the global market, it is seen as one of the most dynamic clusters in the Global South. While initially most firms were run by local entrepreneurs, at present, some of the leading garment exporters in India have set up factories here.

Apart from body building, Karur is a major centre of exports of home furnishings like table cloth, curtains, bed covers and towels. Bhavani and Kumrapalayam are again major centres of production of carpets, both for the domestic and the global markets.

Apart from such modem clusters, there are also traditional artisanal clusters such as Madurai and Kanchipuram that are famous for silk and cotton handloom sarees. Even these clusters have witnessed a degree of modernisation with use of powerlooms in several units.

Question 5.
Write in detail about the types of policies adopted by the Tamil Nadu government to industrialise.
Answer:
The type of policies that are adopted by the Tamil Nadu Government to Industrialise are:- Education, Infrastructure, and Industrial promotion.

(i) Education: Industries require skilled human resources. Therefore, labourers are given technical knowledge apart from basic skills to enrich themselves. Many engineering colleges, polytechnics and Industrial training centres are opened in the country.

(ii) Infrastructure: Excellent infrastructure facilities has contributed to the spread of Industrialisation in smaller towns and villages in the state. Rural electrification, transport and especially minor roads that connect rural parts of the State enabled vast Industrialisation.

(iii) Industrial Promotion: Policies to promote specific sectors like automobile, auto components, bio – technology and information and communication sectors have been formulated to promote Industries of the state.

Question 6.
Explain the role of Entrepreneur?
Answer:
Role of an Entrepreneur:
Entrepreneurs play a most important role in the economic growth and development of a country’s economy.

1. They promote development of industries and help to remove regional disparities by industrialising rural and backward areas.
2. They help the country to increase the GDP and Per Capita Income.
3. They contribute towards the development of society by reducing concentration of income and wealth.
4. They promote capital formation by mobilising the idle savings of the citizens and country’s export trade.
5. Entrepreneurs provide large-scale employment to artisans, technically qualified persons and professionals and work in an environment of changing technology and try to maximise profits by innovations.
6. They enable the people to avail better quality goods at lower prices, which results in the improvement of their standard of living.

VIII. Case studies

Question 1.
Choose a cluster in Tamil Nadu based on online research and write a note on it.
Answer:
An example of a cluster in Tamil Nadu is given below. Students can do online research on their own and write details using this as a sample.

Example case study of a pulp and paper industry in Tamil Nadu.

Tamil Nadu Newprint and papers Limited (http:// www.tnpl.co.in/)

The company: TNPL was formed by the Government of India in 1979 as a public limited company under the provisions of the companies Act of 1956. Objective: The primary objective of the company is to produce newsprint and printing and writing paper using bagasse, a sugarcane residue as the primary raw material.

Assistance: The only paper will in India assisted by the World Bank.

Machinery: The state of art machines were built in flexibility for manufacturing both newsprint and printing and writing papers in the same machine.

Capacity: The latest upgraded capacity enhanced to 2,30,000 tpa in April 2003 through upgradation of the paper machines which is considered as the largest production capacity in India at a single location.

Marketing: The products are being marketed throughout the country and also exported to 20 countries around the world.

Location: The factory is situated at Kagithapuram in Kanir district in Tamil Nadu.

IX. Activity and Project

Question 1.
Write a note on a cluster or a firm near your school/home based on your observations.
Answer:
Here is an sample of a cluster of firm observed in an area is given. This is to help students gain a basic idea about how to undertake this activity. Students can select their location, observe and write findings and do this activity.

Name of the Village: Neikkarapatti, Salem

Industrial cluster: Good quality and high production of Jaggery. The area in around almost all the houses, small or big make it a point to involve themselves in the production of Jaggery.

Type: It is one of the prominent Cottage Industry in the area.

Reason: Most of the farmers in the region cultivate sugarcane and own a jaggery making unit.

Labour: It is a manual jaggery making traditional work, unmindful of the challenges.

Process: The whole process takes about five hours with different names locally as Vellam and Mandai Vellam based on the shape of the mould.

X. Life Skills

Question 1.
Teacher and Students discuss about the entrepreneurs and their activities and Write an a essay in the topic of “If you are like a Entrepreneur”.
Answer:
Note: Students should imagine themselves as entrepreneurs and write an essay on the topic. This is sample essay given as a guideline for the students.

(A stationery shop – business)

Entrepreneurs are business owners. If I am a entrepreneur, first of all, I should start my business in the line of my interest and in part with the demand of the locality and their needs. This would help me to earn profit from it.

I prefer do my entrepreneurial task by staring it at a small level with less financial investment (approx. 10,000) from my uncle who is also an entrepreneur who supports and motivates me.

I purchase my products for sale from the whole saler. The items I purchase are not perishable stationery items. It is easy for me to store it in my house itself. I get bulk orders from nearby shops. Sometimes, latest arrivals of designed erasers, pencil sharpeners will be a attractive one.

As an entrepreneur, I am ready to face risk factors, price hike, less demand, poor quality, seasonal changes and so on. But, I enjoy my work as a businessman and earn profit that gives my career a spark to continue in the same line of business.

Samacheer Kalvi 10th Social Science Industrial Clusters in Tamil Nadu Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
The conversion of raw materials into usable materials is called as
(a) Entrepreneur
(b) Industry
(c) Exports
(d) Clusters
Answer:
(b) Industry

Question 2.
Wind energy is a form of …………… energy.
(a) Solar
(b) Petrol
(c) Hydel
Answer:
(a) Solar

Question 3.
SMESare:
(a) Small and Medium Enterprises
(b) Small, Medium, Export Services
(c) Small and Medium Export Services
(d) Salem Metal Export Services.
Answer:
(a) Small and Medium Enterprises

Question 4.
TIDCo is another government agency to establish ………………. estates.
(a) Industrial
(b) tea
(c) Real
Answer:
(a) Industrial

Question 5.
The Namakkal -Tiruchengode belt in Western Tamil Nadu is known for its ………………. building industry.
(a) Textile
(b) Truck body
(c) Coach
(d) Steel
Answer:
(b) Truck body

Question 6.
MEPZ is a special Economic zone in …………….
(a) Chennai
(b) Trichy
(c) Tirunelveli
Answer:
(a) Chennai

Question 7.
TANS! was formed in the year:
(a) 1945
(b) 1955
(c) 1965
(d) 1975
Answer:
(c) 1965

Question 8.
………………. is an innovator of new ideas in business.
(a) Agriculturalist
(b) Entrepreneur
(c) Both (a) and (b)
(d) None
Answer:
(b) Entrepreneur

Question 9.
There are limits to the ability of agriculture to absorb labour due to the ………………. marginal productivity of land.
(a) Increasing
(b) Decreasing
(c) Steady
(d) None
Answer:
(b) Decreasing

Question 10.
Health care and educational services are diffused across major cities of:
(a) Chennai
(b) Coimbatore
(c) Both (a) and (b)
(d) Tiruppur
Answer:
(c) Both (a) and (b)

Question 11.
Nanguneri SEZ is situated at:
(a) Tiruvallur
(b) Vayalur
(c) Thirunelveli
(d) Tambaram
Answer:
(c) Thirunelveli

Question 12.
………………. is intended to provide low – cost financial support for both setting up of new units and for expansion of existing units.
(a) TIDCO
(b) TANSI
(c) TIIC
(d) MEPZ
Answer:
(c) TIIC

Question 13.
When Government decide to encourage manufacturing using raw material from a region. It may lead to emergence of:
(a) Industrial clusters
(b) More taxation
(c) Subsidy
(d) Both (a) and (b)
Answer:
(a) Industrial clusters

Question 14.
The notion of an “Industrial district” was developed by Prof:
(a) Adam Smith
(b) Marshall
(c) Samuelson
(d) Pigou
Answer:
(b) Marshall

Question 15.
Heavy vechicles factory was set up to manufacture tanks in:
(a) Peramber
(b) Avadi
(c) Tiruchy
(d) Vellore
Answer:
(b) Avadi

Question 16.
Dindigul, Vellore and Amber area are famous for products.
(a) Rice
(b) Wheat
(c) Leather
(d) Marine
Answer:
(c) Leather

Question 17.
The Avadi Industrial Estate was set up in the year:
(a) 1950
(b) 1960
(c) 1970
(d) 1980
Answer:
(a) 1950

Question 18.
IT (Information Technology) Specific Special Economic Zones are located in ………………. locations in Tamil Nadu.
(a) 6
(b) 7
(c) 8
(d) 10
Answer:
(c) 8

Question 19.
The agencies that is formed to promote Industrial growth in the state by setting up Industrial estates are:
(a) SIPCOT
(b) TIDCO
(c) TIIC
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 20.
Over the last ten years a few software firms have moved from Chennai to:
(a) Madurai
(b) Bangalore
(c) Coimbatore
(d) None
Answer:
(c) Coimbatore

II. Fill in the blanks

1. Stand up India scheme was launched in India in the year
2. ………….. is the ability to create and built something.
3. The Industries that produce raw materials for other Industries are called ………….. Industries.
4. If the output is consumed by another producer then, it is called as ………….. goods.
5. Geographical proximity of small and medium enterprise is a chief characteristic of a ………….. cluster.
6. ………….. policies have contributed to the decline of the handloom weaving industry.
7. The Salem steel plant was set up in the year …………..
8. Home Furnishings cluster is located at …………..
9. Integral Coach Factory (ICF) makes …………..
10. MNC means …………..
11. There are ………….. clusters in 13 districts of Tamil Nadu.
12. Transportation and poultry clusters are located in …………..
13. ………….. located in Trichy manufactures Boilers and Turbines.
14. The ………….. district is the top exporter of finished leather goods in the country.
15. SIPGOT was formed in the year …………..
16. MEPZ was established in the year to promote foreign direct investment.
17. A policy was set up in the year ………….. for setting up special Economic zone.
18. TANSI was formed in the year …………..
19. The poison control centre is associated with ………….. SEZ.
20. ………….. is the first Industrial cooperation operating in the domain of small enterprises.
21. There are ………….. export processing zones in the country set up by the Central Government.
22. ………….. schemes provides financial help specifically to SC and ST borrower and woman borrower.
23. Start up India scheme was launched to generate ………….. and creating …………..

Answers:

1. 5^th April 2016
2. Entrepreneur
3. Basic goods
4. capital
5. successful
6. Colonial
7. 1973
8. Karur
9. Railway coaches
10. Multi – National Companies
11. 27
12. Namakkal
13. BHEL
14. Vellore
15. 1971
16. 1984
17. 2000
18. 1965
19. Bio – pharmaceuticals
20. TANSI
21. Seven
22. Standup India
23. employment, wealth

IV. Choose the correct statement

Question 1.
(i) The development of Industries in India can be studied under colonial phase, post independent phase and liberalisation phase.
(ii) Services sector contributes a major share in employment generation.
(iii) Karur is expertised in bus body building and home furnishing.
(iv) Tamil Nadu is the biggest producer of cotton yarn in the country.
(a) (i), (ii) are correct
(b) (iii), (iv) are correct
(c) (i), (iii), (iv) are correct
(d) all are correct
Answer:
(d) all are correct

Question 2.
(i) Only after the economic reforms in 1990 the state has been the entry of hardware and electronic manufacture centres.
(ii) Bhavani and Kumarapalayan are major centres of production of carpets locally and globally.
(iii) Sivakasi region has now become a major centre for printing and fireworks in the country.
(iv) Dindigul and Erode are major employment generators in Knitwear industry.
(a) (i), (ii) are correct
(b) (i), (ii), (iii) are correct
(c) only (ii) and (iv) correct
(d) All are correct
Answer:
(b) (i), (ii), (iii) are correct

V. Pick out odd one

Question 1.
Which is not a export processing zone?
(a) MEPZ
(b) SEZ
(c) TIDCO
(d) ELCOT
Answer:
(c) TIDCO

Question 2.
Tamil Nadu’s Textile clusters does not include
(a) Coimbatore
(b) Padalam
(c) Somanur
(d) Dindigul
Answer:
(d) Dindigul

Question 3.
Which of the following is not under IT cluster?
(a) Nokia
(b) Foxconn
(c) Infosys
(d) MEPZ
Answer:
(d) MEPZ

VI. Write Short Answer

Question 1.
What is the industry?
Answer:
“Any human activity which is engaged in the conversion of raw materials into readily usable materials is called an industry”.

Question 2.
Why is small scale sectors considered important?
Answer:
The small scale sector is seen as important for two reasons.

1. To generate more employment than the large – scale sector.
2. The small scale sectors allows for a larger number of entrepreneurs to emerge from less privileged groups.

Question 3.
What is MEPZ?
Answer:
MEPZ is a special Economic Zone in Chennai and it is one of the seven export processing zones in the country set up by the central government. It was established in 1984. The MPEZ head-quarters is located on GST Road in Tambaram, Chennai.

Question 4.
Write the names of MNCs in Chennai region.
Answer:
MNCs such as Nokia, Foxconn, Samsung and Flextronics.

Question 5.
Write about Bus Body Building Industry clusters.
Answer:
The Namakkal-Tiruchengode belt in western Tamil Nadu is known for its truck body building industry. About 150 of the 250 units in this sector are located in this cluster including 12 large-sized body building houses. Karur is another major hub with more than 50 units. Many entrepreneurs were previous employees in a big firm involved in body building who came out . to set up their own units.

Question 6.
List down the names of the agencies that have played an important role in the Industrialisation of a state.
Answer:

1. SIPCOT – State Industries Promotion Corporation of Tamil Nadu.
2. TANSIDCO – Tamil Nadu Small Industries Development Corporation.
3. TIDCO – Tamil Nadu Industrial Development Corporation
4. TIIC – Tamil Nadu Industrial Investment Corporation Limited.
5. TANSI – Tamil Nadu Small Industries Corporation Limited.

Question 7.
What is meant by SEZ?
Answer:
SEZ means Special Economic Zone in the country with a view to improve free environment for exports.

Question 8.
What do you understand by Stand up India Scheme?
Answer:
It is a scheme launched by the Indian Government with the primary objective of generation employment and wealth creation.

Question 9.
What do you understand by standup India scheme?
Answer:
It is a scheme launched by the Indian Government to facilitate bank loans between ₹ 10 lakh and ₹ 1 crore to atleast one scheduled caste (or) scheduled Tribe borrower and one woman borrower per bank branch for setting up a greenfield enterprise.

Question 10.
How do entrepreneurs promote formation?
Answer:
Entrepreneurs promote capital formation by mobilising the idle savings of the citizens and increasing country’s exports.

VII. Write Brief Answer

Question 1.
Write in detail about the types of industries on the basis of its use, raw ‘ material, ownership, and size.
Answer:
Types of Industries:

On the basis of

Uses:
Consumer goods Industries ⇒ output to the final consumer
Capital goods Industries ⇒ output consumed by another producer.
Basic goods Industries ⇒ output as the raw material for other Industries.

Raw material:
Agricultural sector
Industrial sector
Agro based sector
Leather Industries

Ownership:
Public owned ⇒ Government owned
Private owned ⇒ Private owned Co-operative owned

Size:
Large scale Industries
Small scale Industries
Medium scale Industries
Micro (or) tiny

Question 2.
Name some industrial development agencies and explain them.
Answer:
The following are some agencies that have played a key role in industrialization in the state. SIPCOT: (State Industries Promotion Corporation of Tamil Nadu), 1971 It was formed in the year 1971 to promote industrial growth in the state by setting up industrial estates.

TANSIDCO: (Tamil Nadu Small Industries Development corporation), 1970
TANSIDCO is a state-agency of the state of TN established in the year 1970 to promote small- scale industries in the state. It gives subsidies and provide technical assistance for new firms in the small scale sector.

TIDCO (Tamil Nadu Industrial Development Corporation), 1965:
TIDCO is another government agency to promote industries in the state and to establish industrial estates.

TIIC (Tamil Nadu Industrial Investment Corporation Ltd.), 1949:
TIIC is intended to provide low-cost financial support for both setting up new units and also for expansion of existing units. Though it is meant to meet the requirements of all types of firms, 90% of support goes to micro, small and medium enterprises.

TANSI (Tamil Nadu Small Industries Corporation Ltd.), 1965:
TANSI was formed in 1965 to take over the small scale-units that were set up and run by the Department of Industries and Commerce. It is supposed to be the first industrial corporation operating in the domain for small enterprises.