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Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
If aij = \(\frac{1}{2}\) (3i – 2j) and A = [aij]2×2 is
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 2

Question 2.
What must be the matrix X, if 2X + \(\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right]=\left[\begin{array}{ll}{3} & {8} \\ {7} & {2}\end{array}\right]\) ?
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 3
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 4
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 5

Question 3.
Which one of the following is not true about the matrix \(\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {5}\end{array}\right]\)?
(a) a scalar matrix
(b) a diagonal matrix
(c) an upper triangular matrix
(d) A lower triangular matrix
Solution:
(b) a diagonal matrix

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Question 4.
If A and B are two matrices such that A + B and AB are both defined, then …………
(a) A and B are two matrices not necessarily of same order.
(b) A and B are square matrices of same order.
(c) Number of columns of a is equal to the number of rows of B.
(d) A = B.
Solution:
(b) A and B are square matrices of same order.

Question 5.
If A = \(\left[\begin{array}{rr}{\lambda} & {1} \\ {-1} & {-\lambda}\end{array}\right]\), then for what value of λ, A2 = 0?
(a) 0
(b) ±1
(c) -1
(d) 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 6

Question 6.
If Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 7 and (A + B)2 = A2 + B2, then the values of a and b are ……………….
(a) a = 4, b = 1
(b) a = 1, b = 4
(c) a = 0, b = 4
(d) a = 2, b = 4
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 8
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 9

Question 7.
If Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 10 is a matrix satisfying the equation AAT = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to ………….
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 11
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 12

Question 8.
If A is a square matrix, then which of the following is not symmetric?
(a) A + AT
(b) AAT
(c) ATA
(d) A – AT
Solution:
(b) AAT

Question 9.
If A and B are symmetric matrices of order n, where (A ≠ B), then …………….
(a) A + B is skew-symmetric
(b) A + B is symmetric
(c) A + B is a diagonal matrix
(d) A + B is a zero matrix
Solution:
(b) A + B is symmetric

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Question 10.
If Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 13 and if xy = 1, then det (AAT) is equal to …………..
(a) (a – 1)2
(b) (a2 + 1)2
(c) a2 – 1
(d) (a2 – 1)2
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 14

Question 11.
The value of x, for which the matrix Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 15is singular is ………….
(a) 9
(b) 8
(c) 7
(d) 6
Solution:
(b) Hint: Given A is a singular matrix ⇒ |A| = 0
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 16
⇒ ex-2.e2x+3 – e2+x.e7+x = 0
⇒ e3x+1 – e9+2x = 0
⇒ e3x+1 = e9+2x
⇒ 3x + 1 = 9 + 2x
⇒ 3x – 2x = 9 – 1
⇒ x = 8

Question 12.
If the points (x, -2), (5, 2), (8, 8) are collinear, then x is equal to …………
(a) -3
(b) \(\frac{1}{3}\)
(c) 1
(d) 3
Solution:
(d) Hint: Given that the points are collinear
So, area of the triangle formed by the points = 0
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 17

Question 13.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 18
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 19
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 20

Question 14.
If the square of the matrix Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 21 is the unit matrix of order 2, then α, β and γ should satisfy the relation.
(a) 1 + α2 + βγ = 0
(b) 1 – α2 – βγ = 0
(c) 1 – α2 + βγ = 0
(d) 1 + α2 – βγ = 0
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 22

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Question 15.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 23
(a) Δ
(b) kΔ
(c) 3kΔ
(d) k3Δ
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 24

Question 16.
A root of the equation Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 25 is …………….
(a) 6
(b) 3
(c) 0
(d) -6
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 26

Question 17.
The value of the determinant of Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 27is ……………
(a) -2abc
(b) abc
(c) 0
(d) a2 + b2 + c2
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 28

Question 18.
If x1, x2, x3 as well as y1, y2, y3 are in geometric progression with the same common ratio, then the points (x1, y1), (x2, y2), (x3, y3) are
(a) vertices of an equilateral triangle
(b) vertices of a right-angled triangle
(c) vertices of a right-angled isosceles triangle
(d) collinear
Solution:
(d)

Question 19.
If \(\lfloor.\rfloor\) denotes the greatest integer less than or equal to the real number under consideration and -1 ≤ x < 0, 0 ≤ y < 1, 1 ≤ z ≤ 2, then the value of the determinant Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 29 is …………..
(a) \(\lfloor z\rfloor\)
(b) \(\lfloor y\rfloor\)
(c) \(\lfloor x\rfloor\)
(d) \(\lfloor x\rfloor+ 1\)
Solution:
(a) Hint: From the given values
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 30>

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Question 20.
If a ≠ b, b, c satisfy Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 31 then abc = ……………..
(a) a + b + c
(b) 0
(c) b3
(d) ab + bc
Solution:
(c) Hint: Expanding along R1,
a(b2 – ac) – 2b (3b – 4c) + 2c (3a – 4b) = 0
(b2 – ac) (a – b) = 0
b2 = ac (or) a = b
⇒ abc = b(b2) = b3

Question 21.
If Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 32 then B is given by ………………..
(a) B = 4A
(b) B = -4A
(c) B = -A
(d) B = 6A
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 33
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 34

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Question 22.
If A is skew-symmetric of order n and C ¡s a column matrix of order n × 1, then CT AC is ……………..
(a) an identity matrix of order n
(b) an identity matrix of order 1
(e) a zero matrix of order I
(d) an Identity matrix of order 2
Solution:
(c) a zero matrix of order I
Hint: Given A is of order n × n
C is of order n × 1
so, CT is of order 1 × n
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 35
Let it be equal to (x) say
Taking transpose on either side
(CT, AC)T (x)T .
(i.e.) CT(AT)(C) = x
CT(-A)(C) = x
⇒ CTAC = -x
⇒ x = -x
⇒ 2x = 0
⇒ x = 0

Question 23.
The matrix A satisfying the equation Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 36 is ……………
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 37
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 38

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 39

Question 24.
If A + I = Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 40, then (A + I) (A – I) is equal to …………….
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 41
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 42

Question 25.
Let A and B be two symmetric matrices of the same order. Then which one of the following statements is not true?
(a) A + B ¡s a symmetric matrix
(b) AB ¡s a symmetric matrix
(c) AB = (BA)T
(d) ATB = ABT
Solution:
(b) AB ¡s a symmetric matrix

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5 Read More »

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

Question 1.
Write the following in roster form.
(i) {x ∈ N : x2 < 121 and x is a prime}.
(ii) the set of all positive roots of the equation (x – 1)(x + 1)(x2 – 1) = 0.
(iii) {x ∈ N : 4x + 9 < 52}.
(iv) {x : \(\frac{x-4}{x+2}\) = 3, x ∈ R – {-2}}
Solution:
(i) Let A = { x ∈ N : x2 < 121 and x is a prime }
A = {2, 3, 5, 7}
(ii) The set of positive roots of the equations
(x – 1) (x + 1) (x2 – 1) = 0
(x – 1 ) (x + 1 ) (x + 1) (x – 1) = 0
(x + 1 )2 (x – 1)2 = 0
(x + 1)2 = 0 or (x – 1)2 = 0
x + 1 = 0 or x – 1 = 0
x = -1 or x = 1
A = { 1 }
(iii) Let A = { x ∈ N : 4x + 9 < 52 }
When x = 1, (4) × (1 ) + 9 = 4 + 9 = 13
When x = 2, (4) × (2) + 9 = 8 + 9 = 17
When x = 3, (4) × (3) + 9 = 12 + 9 = 21
When x = 4, (4) × (4) + 9 = 16 + 9 = 25
When x = 5, (4) × (5) + 9 = 20 + 9 = 29
When x = 6, (4) × (6) + 9 = 24 + 9 = 33
When x = 7, (4) × (7) + 9 = 28 + 9 = 37
When x = 8, (4) × (8) + 9 = 32 + 9 = 41
When x = 9, (4) × (9) + 9 = 36 + 9 = 45
When x = 10, (4) × (10) + 9 = 40 + 9 = 49
∴ A = { 1, 2, 3, 4, 5, 6 ,7, 8, 9, 10 }
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 11
(i.e.) x – 4 = 3(x + 2)
x – 4 = 3x + 6
– 4 – 6 = 3x – x
2x = -10 ⇒ x = -5
A = {-5}
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

Question 2.
Write the set {-1, 1} in set builder form.
Solution:
A = {x : x2 – 1 = 0, x ∈ R}

Question 3.
State whether the following sets are finite or infinite.

  1. {x ∈ N : x is an even prime number}
  2. {x ∈ N : x is an odd prime number}
  3. {x ∈ Z : x is even and less than 10}
  4. {x ∈ R : x is a rational number}
  5. {x ∈ N : x is a rational number}

Solution:

  1. Finite set
  2. Infinite set
  3. Infinite
  4. Infinite
  5. Infinite

Question 4.
By taking suitable sets A, B, C, verify the following results:
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(if) A × (B ∪ C) = (A × B) ∪ (A × C).
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B).
(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A).
Solution:
To prove the following results let us take U = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
B = {2, 7, 8, 9}
C = {1, 5, 8, 7}
(i) Let A = {1, 2}, B = {3, 4}, C = {4, 5}
B ∩ C = {3, 4} ∩ {4, 5}
B ∩ C = {4}
A × (B ∩ C) = {1, 2} × {4}
A × (B ∩ C) = { (1,4), (2,4) } —– (1)
A × B = {1, 2} × {3, 4}
A × B = { (1,3), (1, 4), (2, 3), (2, 4)}
A × C = {1, 2} × { 4, 5 }
A × C = {(1, 4), (1, 5), (2, 4), (2, 5)}
(A × B) ∩ (A × C) = {(1, 3), (1, 4), (2, 3), (2, 4)} ∩ { (1, 4), (1, 5), (2, 4), (2, 5)}
(A × B) ∩ (A × C) = {(1, 4), (2, 4)} —- (2)
From equations (1) and (2)
A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To prove A × (B ∪ C) = (A × B) (A × C)
B = {2, 7, 8, 9}, C = {1, 5, 8, 10)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
A × (B ∪ C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)) …. (1)
A × B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9),
(7, 2), (7, 7), (7, 8), (7, 9)}
A × C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∪ (A × C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)} …… (2)
(1) = (2) ⇒ A × (B ∪ C) = (A × B) ∪ (A × C)

(iii) Let A = {1, 2}, B = {2, 3}
A × B = {1, 2} × {2, 3}
A × B = {(1, 2), (1, 3), (2, 2), (2, 3)}
B × A = {2, 3} × {1, 2}
B × A = {(2, 1), (2, 2), (3, 1), (3,2)}
(A × B) ∩ (B × A) = {(1, 2), (1, 3),(2, 2), (2, 3)} ∩ {(2, 1), (2, 2), (3, 1),(3, 2)}
(A × B) ∩ (B × A) = {(2, 2)} ——- (1)
A ∩ B = {1, 2} ∩ {2, 3}
A ∩ B = {2}
B ∩ A = {2, 3} ∩ {1, 2}
B ∩ A = {2}
(A ∩ B) × (B ∩ A) = {2} × {2}
(A ∩ B) × (B ∩ A) = {(2,2)} ———- (2)
From equations (1) and (2)
(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)

(iv) To prove C – (B – A) = (C ∩ A) ∪ (C ∩ B)
B – A = {8, 9}
C = {1, 5, 8, 10}
∴ LHS = C – (B – A) = {1, 5, 10} …… (1)
C ∩ A = {1}
U = {1, 2, 5, 7, 8, 9, 10}
B = {2, 7, 8, 9} ∴ B’ = {1, 5, 10}
C ∩ B = {1, 5, 10}
R.H.S. (C ∩ A) ∪ (C ∩ B) = {1} ∪ {1, 5, 10}
= {1, 5, 10} ……. (2)
(1) = (2) ⇒ LHS = RHS

(v) Let A = {1, 2, 3, 4} , B = {3, 4, 5, 6}, C = { 5, 6, 7, 8 )
B – A = {3, 4, 5, 6} – {1, 2, 3, 4}
B – A = {5, 6}
(B – A) ∩ C = {5, 6} ∩ {5, 6, 7, 8}
(B – A) ∩ C = {5, 6} ——– (1)
(B ∩ C) = {3, 4, 5, 6} ∩ {5, 6, 7, 8}
B ∩ C = {5, 6}
(B ∩ C) – A = {5, 6} – {1,2,3,4}
(B ∩ C) – A = {5, 6} ——- (2)
C – A = {5, 6, 7, 8} – {1, 2, 3, 4}
C – A = {5, 6, 7, 8}
B ∩ (C – A) = {3, 4, 5, 6} ∩ {5, 6, 7, 8}
B ∩ (C – A) = {5, 6} ——– (3)
From equations (1) , (2) and (3)
(B – A) ∩ C = (B ∩ C) – A = B ∩(C – A)

(vi) To prove (B – A) ∪ C ={1, 5, 8, 9, 10}
B – A = {8, 9},
C = {1, 5, 8, 10}
(B – A) ∪ C = {1, 5, 8, 9, 10} ……. (1)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A – C = {2, 7}
(B ∪ C) – (A – C) = {1, 5, 8, 9, 10} ……… (2)
(1) = (2)
⇒ (B – A) ∪ C = (B ∪ C) – (A – C)

Question 5.
Justify the trueness of the statement.
“An element of a set can never be a subset of itself.”
Solution:
“An element of a set can never be a subset of itself ”
The statement is correct
Let A = {a, b, c, d} for a ∈ A
‘a’ cannot be a subset of ‘a’

Question 6.
If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).
Solution:
n(P( A)) = 1024 = 210 ⇒ n( A) = 10
n(A ∪ B) = 15
n(P(B)) = 32 = 25 ⇒ n(B) = 5
We know n(A ∪ B) = n{A) + n(B) – n(A ∩ B)
(i.e.) 15 = 10 + 5 – n(A ∩ B)
⇒ n(A ∩ B) = 15 – 15 = 0

Question 7.
If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A(A ∆ B)).
Solution:
Given n(A ∩ B) = 3 and n(A ∪ B) = 10
A ∆ B = (A – B) ∪ (B – A)
n(A ∆ B) = n [ (A – B ) ∪ (B – A)]
n(A ∆ B) = n(A – B) + n(B – A) —— (1)
(Since A – B and B – A are disjoint sets)
A ∪ B = (A – B) ∪ (B – A) ∪ (A ∩ B)
n(A ∪ B) = n[(A – B) ∪ (B – A) ∪ (A ∩ B)]
n(A ∪ B) = n (A – B) + n (B – A) + n (A ∩ B)
(Since A – B, B – A and A ∩ B are disjoint sets)
n(A ∪ B) = n(A ∆ B) + n(A ∩ B)
10 = n(A ∆ B) + 3
n(A ∆ B) = 10 – 3 = 7
∴ n(P(A ∆ B)) = 27 = 128

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

Question 8.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.
Solution:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 9.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.
Solution:
Given A and B be two sets such that n (A) = 3 and n(B) = 2.
Also given (x, 1), (y, 2), (z, 1) ∈ A × B
A = { x, y, z }, B = {1, 2}

Question 10.
If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (-1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.
Solution:
n(A × A) = 16 ⇒ n( A) = 4
S ={(-1, 0), (-1, 1), (0, 2), (1, 2)}

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 Additional Questions

Question 1.
Write the following sets in roster form
(a) {x ∈ N; x3 < 1000}
(b) {The set of positive roots of the equation (x2 – 4) (x3 – 27) = 0}
Solution:
(a) A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) B = {2, 3}

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

Question 2.
By taking suitable sets A, B, C verify the following results
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) (B – A) ∪ C = (B ∪ C) – (A – C)
Solution:
Prove by yourself

Question 3.
Given n(A) = 7; n(B) = 8 and n(A ∪ B) = 10 find n[P(A ∩ B)].
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i.e.,) 10 = 7 + 8 – n(A ∩ B)
⇒ n(A ∩ B) = 7 + 8 – 10 = 5
So n[P(A ∩ B)] = 25 = 32

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Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics

Samacheer Kalvi 10th Science Nuclear Physics Textual Solved Problems

Question 1.
Identify A, B, C, and D from the following nuclear reactions.
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 1
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 2
A is alpha particle, B is neutron, C is proton and D is electron.

Question 2.
A radon specimen emits radiation of 3.7 × 103 GBq per second. Convert this disintegration in terms of a curie, (one curie = 3.7 × 1010 disintegration per second)
Solution:
1 Bq = one disintegration per second
one curie = 3.7 × 1010 Bq
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 3

Question 3.
\(_{92} \mathrm{U}^{235}\) experiences one α – decay and one β – decay. Find the number of neutrons in the final daughter nucleus that is formed.
Solution:
Let X and Y be the resulting nucleus after the emission of the alpha and beta particles respectively.
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 4
Number of neutrons = Mass number – Atomic number = 231 – 91 = 140.

Question 4.
Calculate, the amount of energy released when a radioactive substance undergoes fusion and results in a mass defect of 2 kg.
Solution:
Mass defect in the reaction (m) = 2 kg
Velocity of light (c) = 3 × 108 ms-1
By Einstein’s equation,
Energy released E = mc2
= 2 × (3 × 108)2
= 1.8 × 1017 J.

Samacheer Kalvi 10th Science Nuclear Physics Textual Evaluation

I. Choose the correct answer

Question 1.
Man – made radioactivity is also known as _____.
(a) Induced radioactivity
(b) Spontaneous radioactivity
(c) Artificial radioactivity
(d) (a) & (c).
Answer:
(d) (a) & (c).

Question 2.
Unit of radioactivity is:
(a) roentgen
(b) curie
(c) becquerel
(d) all the above
Answer:
(d) all the above

Question 3.
Artificial radioactivity was discovered by _____.
(a) Becquerel
(b) Irene Curie
(c) Roentgen
(d) Neils Bohr.
Answer:
(b) Irene Curie

Question 4.
In which of the following, no change in mass number of the daughter nuclei takes place:
(i) a decay;
(ii) P decay
(iii) y decay
(iv) neutron decay
(a) (i) is correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Answer:
(b) (ii) and (iii) are correct

Question 5.
_____ isotope is used for the treatment of cancer.
(a) Radio Iodine
(b) Radio Cobalt
(c) Radio Carbon
(d) Radio Nickel.
Answer:
(b) Radio Cobalt

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Question 6.
Gamma radiations are dangerous because:
(a) it affects eyes and bones
(b) it affects tissues
(c) it produces genetic disorder
(d) it produces an enormous amount of heat
Answer:
(c) it produces genetic disorder

Question 7.
_____ aprons are used to protect us from gamma radiations.
(a) Lead oxide
(b) Iron
(c) Lead
(d) Aluminium.
Answer:
(c) Lead

Question 8.
Which of the following statements is / are correct?
(i) α particles are photons
(ii) Penetrating power of γ radiation is very low
(iii) Ionization power is maximum for α rays
(iv) Penetrating power of γ radiation is very high
(a) (i) & (ii) are correct
(b) (ii) & (iii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct.
Answer:
(d) (iii) & (iv) are correct.

Question 9.
Proton-Proton chain reaction is an example of:
(a) Nuclear fission
(b) α – decay
(c) Nuclear fusion
(d) β – decay
Answer:
(c) Nuclear fusion

Question 10.
In the nuclear reaction \(_6^{\mathrm{X}^{12}} \stackrel{\alpha \text { decay }}{\longrightarrow} \mathrm{z}^{\mathrm{Y}^{\mathrm{A}}}\), the value of A & Z.
(a) 8, 6
(b) 8, 4
(c) 4, 8
(d) cannot be determined with the given data.
Answer:
(c) 4, 8

Question 11.
Kamini reactor is located at _____.
(a) Kalpakkam
(b) Koodankulam
(c) Mumbai
(d) Rajasthan.
Answer:
(a) Kalpakkam

Question 12.
Which of the following is/are correct?
(i) Chain reaction takes place in a nuclear reactor and an atomic bomb.
(ii) The chain reaction in a nuclear reactor is controlled.
(iii) The chain reaction in a nuclear reactor is not controlled.
(iv) No chain reaction takes place in an atom bomb.
(a) (i) only correct
(b) (i) & (ii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct
Answer:
(b) (i) & (ii) are correct

II. Fill in the blanks

Question 1.
One roentgen is equal to ______ disintegrations per second?
Answer:
3.7 × 1010.

Question 2.
Positron is an _____.
Answer:
antiparticle of electron.

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Question 3.
Anaemia can be cured by _____ isotope.
Answer:
Radio iron (Fe59).

Question 4.
Abbreviation of ICRP _____.
Answer:
International Commission on Radiological Protection.

Question 5.
_____ is used to measure the exposure rate of radiation in humans.
Answer:
Roentgen.

Question 6.
_____ has the greatest penetration power.
Answer:
Gamma ray.

Question 7.
\(z^{\mathrm{Y}^{\mathrm{A}}} \rightarrow_{\mathrm{Z}+1} \mathrm{Y}^{\mathrm{A}}+\mathrm{X}\); Then X is _____.
Answer:
\(_{-1} \mathrm{e}^{0}\) (β decay).

Question 8.
\(z^{\mathrm{X}^{\mathrm{A}}} \rightarrow_{\mathrm{Z}}^{\mathrm{Y}^{\mathrm{A}}}\) This reaction is possible in _____ decay.
Answer:
Gamma (γ).

Question 9.
The average energy released in each fusion reaction is about _____ J.
Answer:
3.84 × 10-12.

Question 10.
Nuclear fusion is possible only at an extremely high temperature of the order of _____ K.
Answer:
107 to 109.

Question 11.
The radioisotope of _____ helps to increase the productivity of crops.
Answer:
phosphorous (P – 32).

Question 12.
If radiation exposure is 100 R, it may cause _____.
Answer:
fatal disease.

III. State whether the following statements are true or false: If false, correct the statement

Question 1.
Plutonium -239 is a fissionable material.
Answer:
True.

Question 2.
Elements having an atomic number greater than 83 can undergo nuclear fusion.
Answer:
False.
Correct Statement: Elements having an atomic number greater than 83 can undergo nuclear fusion.

Question 3.
Nuclear fusion is more dangerous than nuclear fission.
Answer:
False.
Correct Statement: Nuclear fission is more dangerous than nuclear fusion. Because the average energy released in fission (3.2 × 10-11 J) process is more than the average energy released in fusion (3.84 × 10-12 J).

Question 4.
Natural uranium U-238 is the core fuel used in a nuclear reactor.
Answer:
False.
Correct Statement: U-238 is not a fissile material but are abundant in nature. But in a reactor, this can be converted into a fissile material Pu239 and U233. Only fissile materials are used in the fuel of a nuclear reactor.

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Question 5.
If a moderator is not present, then a nuclear reactor will behave like an atom bomb.
Answer:
True.

Question 6.
During one nuclear fission on an average, 2 to 3 neutrons are produced.
Answer:
True.

Question 7.
Einstein’s theory of mass-energy equivalence is used in nuclear fission and fusion.
Answer:
True.

IV. Match the following

Question 1.

1. BARC (a) Kalpakkam
2. India’s first atomic power station (b) Apsara
3. IGCAR (c) Mumbai
4. The first nuclear reactor in India (d) Tarapur

Answer:
1. (c) Mumbai
2. (d) Tarapur
3. (a) Kalpakkam
4. (b) Apsara

Question 2.

1. Fuel (a) lead
2. Moderator (b) heavy water
3. Coolant (c) Graphite
4. Shield (d) Uranium

Answer:
1. (d) uranium
2. (c) Graphite
3. (b) heavy water
4. (a) lead

Question 3.

1. Soddy Fagan (a) Natural radioactivity
2. Irene Curie (b) Displacement law
3. Henry Becquerel (c) Mass energy equivalence
4. Albert Einstein (d) Artificial Radioactivity

Answer:
1. (b) Displacement law
2. (d) Artificial Radioactivity
3. (a) Natural radioactivity
4. (c) Mass energy equivalence

Question 4.

1. Uncontrolled fission Reaction (a) Hydrogen Bomb
2. Fertile material (b) Nuclear Reactor
3. Controlled fission Reaction (c) Breeder reactor
4. Fusion reaction (d) Atom bomb

Answer:
1. (d) Atom bomb
2. (c) Breeder reactor
3. (b) Nuclear Reactor
4. (a) Hydrogen Bomb

Question 5.

1. Co – 60 (a) Age of fossil
2. I – 13 (b) Function of Heart
3. Na – 24 (c) Leukaemia
4. C – 14 (d) Thyroid disease

Answer:
1. (c) Leukemia
2. (d) Thyroid disease
3. (b) Function of Heart
4. (a) Age of fossil

V. Arrange the following in the correct sequence

Question 1.
Arrange in descending order, on the basis of their penetration power.

  1. Alpha rays
  2. Beta rays
  3. Gamma rays
  4. Cosmic rays.

Answer:

  1. Gamma rays
  2. Beta rays
  3. Alpha rays
  4. Cosmic rays.

Question 2.
Arrange the following in the chronological order of discovery.

  1. A nuclear reactor
  2. Radioactivity
  3. Artificial radioactivity
  4. Discovery of radium.

Answer:

  1. Radioactivity (1896)
  2. Discovery of radium (1898)
  3. Artificial radioactivity (1934)
  4. Nuclear reactor (1942).

VI. Use the analogy to fill in the blank

Question 1.
Spontaneous process : Natural Radioactivity, Induced process: _____.
Answer:
Artificial radioactivity
(or)
Man – made activity.

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Question 2.
Nuclear Fusion : Extreme temperature, Nuclear Fission: _____.
Answer:
Room temperature.

Question 3.
Increasing crops : Radio phosphorous, Effective functioning of heart: _____.
Answer:
Radio sodium (Na24).

Question 4.
Deflected by electric field : α ray, Null Deflection: _____.
Answer:
γ ray (Gamma – ray).

VII. Numerical Problems

Question 1.
\(\mathrm{8} \mathrm{8}^{\mathrm{Ra}^{226}}\) experiences three α-decay. Find the number of neutrons in the daughter element.
Solution:
\(\mathrm{8} \mathrm{8}^{\mathrm{Ra}^{226}}\) consider as a parent element that is \(\mathrm{8} \mathrm{8}^{\mathrm{X}^{226}}\) and their daughter element is \(z^{\mathrm{Y}^{\mathrm{A}}}\)
According to α decay process,
\(88^{\mathrm{X} 26} \stackrel{3 \alpha \text { decay }}{\longrightarrow} 82^{214}+3 \alpha\) decay
During the 3α decay, the atomic number decreases by 6 and mass number decreases by 12.
So the number of neutrons in the daughter element
N = A – Z
N = 214 – 88 = 126
Number of neutrons in the daughter element N = 126.

Question 2.
A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration in to becquerel (one curie = 3.7 × 1010 Bq).
Solution:
Cobalt specimen emits induced radiation = 75.6 millicurie per second
(1 curie = 3.7 × 1010 Bq)
So 75.6 millicurie = 75.6 × 103 × 1 curie
= 75.6 × 10-3 × 3.7 × 1010 Bq
= 279.72 × 107
= 2.7972 × 109 Bq
75.6 millicurie per second is equivalent to 2.7972 × 109 Bq.

VIII. Assertion and Reason Type Questions

Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

Question 1.
Assertion: A neutron impinging on U235, splits it to produce Barium and Krypton.
Reason: U-235 is a fissile material.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion: In a β – decay, the neutron number decreases by one.
Reason: In β – decay atomic number increases by one.
Answer:
(d) The assertion is false, but the reason is true.
Explanation: In β – decay there is no change in the mass number of the daughter nucleus but the atomic number increases by one.

Question 3.
Assertion: Extreme temperature is necessary to execute nuclear fusion.
Reason: In nuclear fusion, the nuclei of the reactants combine releasing high energy.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 4.
Assertion: Control rods are known as ‘Neutron seeking rods’
Reason: Control rods are used to perform a sustained nuclear fission reaction.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Control rods are used to control the number of neutrons in order to have a sustained the chain reaction. They absorb the neutrons, (they seeking the neutrons)

IX. Answer in one or two words (VSA)

Question 1.
Who discovered natural radioactivity?
Answer:
Henri Becquerel was discovered natural radioactivity.

Question 2.
Which radioactive material is present in the ore of pitchblende?
Answer:
Uranium

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Question 3.
Write any two elements which are used for inducing radioactivity?
Answer:

  1. Boron and Aluminium.
  2. Alpha particle and neutron.

Question 4.
Write the name of the electromagnetic radiation which is emitted during a natural radioactivity.
Answer:
Gamma rays

Question 5.
If A is a radioactive element which emits an α-particle and produces \({ { _{ 104 }{ Rf } } }^{ 259 }\). Write the atomic number and mass number of the element A.
Answer:
In α decay
\(\begin{array}{l}{_{\mathrm{z}} \mathrm{X}^{\mathrm{A}} \frac{\alpha \text { decay }}{263} \times \mathrm{z}-2 \mathrm{Y}^{\mathrm{A}-4}+_{2} \mathrm{He}^{4}(\alpha \text { decay })} \\ {106^{\mathrm{X}^{263}} \stackrel{\alpha \text { decay }}{\longrightarrow}_{104} \mathrm{Rf}^{259}+_{2} \mathrm{He}^{4}}\end{array}\)
In element A having atomic number is 106 and mass number is 263.

Question 6.
What is the average energy released from a single fission process?
Answer:
The average energy released from a single fission process is about 3.2 × 10-11 J.

Question 7.
Which hazardous radiation is the cause for the genetic disorders (or) effect?
Answer:
Radioactive radiations

Question 8.
What is the amount of radiation that may cause the death of a person when exposed to it?
Answer:
When the body is exposed to about 600 R, it leads to death.

Question 9.
When and where was the first nuclear reactor built?
Answer:
The first nuclear reactor was built in 1942 in Chicago, USA.

Question 10.
Give the SI unit of radioactivity.
Answer:
Becquerel

Question 11.
Which material protects us from radiation?
Answer:
Lead coated aprons and lead gloves should be used while working with the hazardous area. These materials are used to protects us from radiation.

X. Answer the following questions in a few sentences.

Question 1.
Write any three features of natural and artificial radioactivity.
Answer:

Natural radioactivity Artificial radioactivity
1. Emission of radiation due to the self-disintegration of a nucleus. 1. Emission of radiation due to the disintegration of a nucleus through the induced process.
2. Alpha, Beta and Gamma radiations are emitted. 2. Mostly elementary particles such as neutron, positron, etc. are emitted.
3. It is a spontaneous process. 3. It is an induced process.

Question 2.
Define critical mass.
Answer:
The minimum mass of fissile material necessary to sustain the chain reaction is called ‘critical mass (mc). It depends on the nature, density and the size of the fissile material.

Question 3.
Define One roentgen.
Answer:
One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.

Question 4.
State Soddy and Fagan’s displacement law.
Answer:
During a radioactive disintegration, the nucleus which undergoes disintegration is called a parent nucleus and that which remains after the disintegration is called the daughter nucleus.

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Question 5.
Give the function of control rods in a nuclear reactor.
Answer:
Control rods are used to control the number of neutrons in order to have sustained chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.

Question 6.
In Japan, some of the newborn children are having congenital diseases. Why?
Answer:
During the Second World War American, a bomber dropped the nuclear weapons over the Japanese cities of Hiroshima and Nagasaki. In the explosion of the atomic bomb to release the high energy dangerous radiation. In the explosion period, Japanese peoples are affected by radiation. This is the reason in Japan, some of the newborn children are having congenital diseases.

Question 7.
Mr Ramu is working as an X – ray technician in a hospital. But, he does not Wear the lead aprons. What suggestion will you give to Mr Ramu?
Answer:
X – rays have a destructive effect on living tissue. When the human body is exposed to X – rays, it causes redness of the skin, sores and serious injuries to the tissues and glands. They destroy the white corpuscles of the blood. If you don’t wear the lead aprons these kinds of diseases formed in your body. In my suggestion, you must wear lead aprons.

Question 8.
What is stellar energy?
Answer:
Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as stellar energy.

Question 9.
Give any two uses of radioisotopes in the field of agriculture?
Answer:

  • The radioisotope of phosphorus (P – 32) helps to increase the productivity of crops.
  • The radiations from the radioisotopes can be used to kill the insects and parasites and prevent the wastage of agricultural products.

XI. Answer the following questions in detail.

Question 1.
Explain the process of controlled and uncontrolled chain reactions.
Answer:
(a) Controlled chain reaction

  • In the controlled chain reaction, the number of neutrons released is maintained to be one. This is achieved by absorbing the extra neutrons with a neutron absorber leaving only one neutron to produce further fission.
  • Thus, the reaction is sustained in a controlled manner. The energy released due to a controlled chain reaction can be utilized for constructive purposes.
  • The controlled chain reaction is used in a nuclear reactor to produce energy in a sustained and controlled manner.

(b) Uncontrolled chain reaction:

  • In the uncontrolled chain reaction, the number of neutrons multiplies indefinitely and causes fission in a large amount of the fissile material.
  • This results in the release of a huge amount of energy within a fraction of a second.
  • This kind of chain reaction is used in the atom bomb to produce an explosion.
    Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 5

Question 2.
Compare the properties of Alpha, Beta and Gamma radiations.
Answer:

Properties α rays β rays γ rays
What are they? Helium nucleus \(\left(_{2} \mathrm{He}^{4}\right)\) consisting of two protons and two neutrons. They are electrons \(\left(_{-1} e^{\mathrm{0}}\right)\), basic elementary particle in all atoms. They are electromagnetic waves consisting of photons.
Charge Positively charged particles. Charge of each alpha particle = +2e Negatively charged particles. Charge of each beta particle = -e Neutral particles. Charge of each gamma particle = zero
Ionising Power 100 time greater than β rays and 10,000 times greater than γ rays Comparatively low Very less ionization power
Penetrating power Low penetrating power (even stopped by a thick paper) Penetrating power is greater than that of α rays. They can penetrate through a thin metal foil. They have a very high penetrating power greater than that of β rays. They can penetrate through thick metal blocks.
Effect of an electric and magnetic field Deflected by both the fields. (in accordance with Fleming’s left-hand rule) Deflected by both the fields, but the direction of deflection is opposite to that for alpha rays. (in accordance with Fleming’s left-hand rule) They are not deflected by both the fields.
Speed Their speed ranges from 1/10 to 1/20 times the speed of light. Their speed can go up to 9/10 times the speed of light. They travel with the speed of light.

Question 3.
What is a nuclear reactor? Explain its essential parts with their functions.
Answer:
Nuclear reactor: A Nuclear reactor is a device in which the nuclear fission reaction takes place in a self – sustained and controlled manner to produce electricity.
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 6
Components of a Nuclear Reactor:
The essential components of a nuclear reactor are

  • Fuel: A fissile material is used as the fuel. The commonly used fuel material is uranium.
  • Moderator: A moderator is used to slow down the high energy neutrons to provide slow neutrons. Graphite and heavy water are commonly used moderators.
  • Control rod: Control rods are used to control the number of neutrons in order to have a sustained a chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.
  • Coolant: A coolant is used to remove the heat produced in the reactor core, to produce steam. This steam is used to run a turbine in order to produce electricity. Water, air and helium are some of the coolants.
  • Protection wall: A thick concrete lead wall is built around the nuclear reactor in order to prevent the harmful radiations from escaping into the environment.

XII. HOT Questions

Question 1.
Mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta decay that can occur.
Answer:
Mass number A = 232
Atomic number Z = 90
Daughter element:
Mass number A = 208
Atomic number Z = 82
Difference in mass number = 232 – 208 = 24
Difference in atomic number
= 90 – 82 = 8
Atomic number of α = 2
Atomic number of β = -1
Mass number of α = 4
Mass number of β = 0
Difference in mass number in transformations
= 24
Number of a decays = \(\frac{24}{4}\) = 6
Difference in atomic number = 8
ΔZ = 6α + 4β
= 6(2) + 4(-1)
= 12 – 4
= 8
∴ Number of β decays = 4
∴ Number of α decays = 6
∴ Number of β decays = 4

Question 2.
‘X – rays should not be taken often’. Give the reason.
Answer:

  • Radiation does involve in X – rays tests and isotope scans (in nuclear medicine) are too low to cause immediate hazardous effects.
  • If should be taken often, X – ray radiation from medical examinations though slightly increases one’s risk for cancer which can occur year or decades after X-ray exposure.

Question 3.
Cell phone towers should be placed far away from the residential area. why?
Answer:

  1. Living near a cell phone tower is not healthy. There is multiple health risks associated with living near a cell phone tower.
  2. Cell phone towers communicate by use pulsed microwave signals (radiofrequency radiation) with each other.
  3. That is the reason cell phone towers should be placed far away from the residential area.

Samacheer Kalvi 10th Science Nuclear Physics Additional Questions

I. Choose the best Answer.

Question 1.
Radium was discovered by _____.
(a) Marie curie
(b) Irene curie
(c) Henri Becquerel
(d) F. Joliot.
Answer:
(a) Marie Curie

Question 2.
How many radioactive substances discovered so far?
(a) 83
(b) 92
(c) 43
(d) 29
Answer:
(d) 29

Question 3.
The SI unit of Radioactivity is _____.
(a) Curie
(b) Rutherford
(c) Becquerel
(d) Roentgen (R).
Answer:
(c) Becquerel

Question 4.
Radioactivity is _____.
(a) increases with increase in temperature
(b) increases with increase in pressure
(c) depends on the number of electrons
(d) purely a nuclear phenomenon.
Answer:
(d) purely a nuclear phenomenon

Question 5.
Which of the following processes is a spontaneous process?
(a) Artifical radioactivity
(b) Natural radioactivity
(c) Photoelectric effect
(d) Collisions
Answer:
(b) Natural radioactivity

Question 6.
The charge of the β rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
Answer:
(c) -e

Question 7.
The charge of the γ rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
Answer:
(b) 0

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Question 8.
The atomic number of the elements that exhibit artifical radioactivity is:
(a) more than 82
(b) more than 83
(c) less than 83
(d) less than 82
Answer:
(c) less than 83

Question 9.
Arrange α, β, γ rays in the increasing order of their ionizing power.
(a) α, β, γ
(b) β, α, γ
(c) γ, β, α
(d) γ, α, β.
Answer:
(c) γ, β, α

Question 10.
Which produces a charge of 2.58 × 10-4 Coulomb in 1 Kg of air?
(a) Curie
(b) Becquerel
(c) Rutherford
(d) Roentgen
Answer:
(d) Roentgen

Question 11.
Ionising power of the γ rays _____.
(a) Comparatively very high ionization power
(b) 100 times greater than the α rays
(c) 100 times greater than the β rays
(d) Comparatively very less ionization power.
Answer:
(d) Comparatively very less ionization power.

Question 12.
Ionization power maximum for _____.
(a) neutrons
(b) α particles
(c) γ rays
(d) β particles.
Answer:
(b) α particles

Question 13.
Charge of gamma particle is:
(a) +2e
(b) -e
(c) Zero
(d) +1e
Answer:
(c) Zero

Question 14.
Which has low penetrating power?
(a) α rays
(b) γ rays
(c) β rays
(d) X rays.
Answer:
(a) α rays

Question 15.
In β – decay _____.
(a) atomic number decreases by one
(b) the mass number decreases by one
(c) proton number remains the same
(d) neutron number decreases by one.
Answer:
(d) neutron number decreases by one

Question 16.
In which decay the energy level of the nucleus changes:
(a) α – decay
(b) β – decay
(c) γ – decay
(d) neutron decay
Answer:
(c) γ – decay

Question 17.
In γ – decay _____.
(a) atomic number decreases by one
(b) there is no change in atomic and mass number
(c) energy only changes in the decay process
(d) both (b) and (c).
Answer:
(d) both (b) and (c).

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Question 18.
The unit of decay constant is _____.
(a) no unit
(b) second
(c) second-1
(d) curie.
Answer:
(c) second-1

Question 19.
The range of temperature required for nuclear fusion is from:
(a) 107 to 109 K
(b) 10-9 to 10-7 K
(c) 105 to 109
(d) 105 to 107 K
Answer:
(a) 107 to 109 K

Question 20.
1 Rd is equal to _____.
(a) 106 decay / second
(b) 1 decay / second
(c) 3.7 × 1010 becquerel
(d) 1.6 × 1012 decay / second.
Answer:
(a) 106 decay / second

Question 21.
An element \(Z^{X^{A}}\) successively undergoes three α decays and four β decays and gets converted an element Y are respectively _____.
(a) \({ { _{ Z-6 }{ Y } } }^{ A-12 }\)
(b) \({ { _{ Z+2 }{ Y } } }^{ A-12 }\)
(c) \({ { _{ Z-2 }{ Y } } }^{ A-12 }\)
(d) \({ { _{ Z-10 }{ Y } } }^{ A-12 }\).
Answer:
(c) \({ { _{ Z-2 }{ Y } } }^{ A-12 }\)

Question 22.
In the nuclear reaction 88Ra226 → X + 2He4 X is:
(a) 90Th234
(b) 91Pa234
(c) 86Rn222
(d) 88Rn226
Answer:
(d) 88Rn226

Question 23.
Which one of the following is used in the treatment of skin diseases _____.
(a) Na24
(b) I31
(c) Fe59
(d) P32.
Answer:
(d) P32.

Question 24.
Anaemia can be diagnosed by _____.
(a) \({ { _{ 15 }{ P } } }^{ 31 }\)
(b) \({ { _{ 15 }{ P } } }^{ 32 }\)
(c) \({ { _{ 26 }{ P } } }^{ 59 }\)
(d) \({ { _{ 11 }{ P } } }^{ 24 }\).
Answer:
(c) \({ { _{ 26 }{ P } } }^{ 59 }\)

Question 25.
Which is used as a coolant?
(a) Graphite
(b) Liquid sodium
(c) Boron
(d) Cadmium
Answer:
(b) Liquid sodium

Question 26.
The energy released per fission is _____.
(a) 220 MeV
(b) 300 MeV
(c) 250 MeV
(d) 200 MeV.
Answer:
(d) 200 MeV.

Question 27.
In the reaction 1N14 + 0n1 → X + 1H1 X is:
(a) 15P30
(b) 6C14
(c) 6C12
(d) 11Na23
Answer:
(c) 6C12

Question 28.
Natural uranium consists of _____.
(a) 99.72 % of U-238
(b) 0.28 % of U-238
(c) 0.72 % of U-238
(d) 99.28 % of U-238.
Answer:
(d) 99.28 % of U-238.

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Question 29.
The number of power reactors in India is _____.
(a) 14
(b) 12
(c) 7
(d) 2.
Answer:
(a) 14

Question 30.
In the nucleus of 11Na23 the number of protons and neutrons are:
(a) 12, 11
(b) 10, 12
(c) 11, 12
(d) 11, 23
Answer:
(c) 11, 12

Question 31.
The moderator used in nuclear reactor is _____.
(a) cadmium
(b) boron carbide
(c) heavy water
(d) uranium \(\left(_{92} \mathrm{U}^{235}\right)\).
Answer:
(c) heavy water

Question 32.
The first nuclear reactor was built at _____.
(a) Kalpakkam, India
(b) Hiroshima, Japan
(c) Chicago, USA
(d) Trombay, Bombay.
Answer:
(c) Chicago, USA

Question 33.
Which of the following is used in the treatment of skin cancer?
(a) Radio Cobalt
(b) Radio gold
(c) Radio Cobalt and radio gold
(d) none of the above
Answer:
(c) Radio Cobalt and radio gold

Question 34.
The explosion of an atom bomb is based on the principle of _____.
(a) uncontrolled fission reaction
(b) fusion reaction
(c) controlled fission reaction
(d) none of the above.
Answer:
(a) uncontrolled fission reaction

Question 35.
The reactor in which no moderator used is _____.
(a) fast breeder reactor
(b) pressurised water reactor
(c) pressurised heavy water reactor
(d) boiled water reactor.
Answer:
(a) fast breeder reactor

Question 36.
The number of neutrons present in 92U235 is:
(a) 133
(b) 143
(c) 43
(d) 243
Answer:
(b) 143

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Question 37.
In fast breeder, the coolant system used is _____.
(a) heavy water
(b) light water
(c) liquid sodium
(d) boiled water.
Answer:
(c) liquid sodium

Question 38.
The only reactor in the world which uses U-233 as fuel is _____.
(a) Zerlina
(b) Purnima
(c) Kamini
(d) Tires.
Answer:
(c) Kamini

Question 39.
The temperature of the interior of Sun is about _____.
(a) 1.4 × 107 K
(b) 108 K
(C) 14 × 107 K
(d) 600 K.
Answer:
(a) 1.4 × 107 K

Question 40.
Total energy radiated by Sun is about _____.
(a) 3.6 × 1028 Js-1
(b) 3.8 × 1028 Js-1
(c) 3.8 × 1026 Js-1
(d) 3.8 × 1023 Js-1.
Answer:
(c) 3.8 × 1026 Js-1

II. Fill in the blanks

Question 1.
Cathode rays are discovered by _____.
Answer:
J.J. Thomson.

Question 2.
Positive rays discovered by _____.
Answer:
Goldstein.

Question 3.
The chargeless particles are called neutron, it was discovered by _____.
Answer:
James Chadwick.

Question 4.
Ernest Rutherford explained that the mass of an atom is concentrated in its central part called _____.
Answer:
Nucleus.

Question 5.
The radioactive elements emit harmful radiations are ____, ____, ____ rays.
Answer:
alpha, beta, gamma.

Question 6.
_____ is an spontaneous process.
Answer:
Natural radioactivity.

Question 7.
The element whose atomic number is more than 83 undergoes _____.
Answer:
spontaneous process.

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Question 8.
______ radioactive material is present in the ore of pitchblende.
Answer:
Uranium.

Question 9.
_____ are the example of artificial (or) man-made radioactive elements.
Answer:
Boron, Aluminium.

Question 10.
The element whose atomic number is less than 83 undergoes _____.
Answer:
induced radioactivity.

Question 11.
______ is an controlled manner.
Answer:
Artificial radioactivity.

Question 12.
Spontaneous radioactivity is also known as _____.
Answer:
Natural radioactivity.

Question 13.
One Curie is equal to _____ disintegrations per second.
Answer:
3.7 × 1010

Question 14.
One Rutherford (Rd) is equal to ______ disintegrations per second.
Answer:
106

Question 15.
The radioactive displacement law is framed by _____.
Answer:
Soddy and Fajan.

Question 16.
During the α decay process, the atomic number is ______ by 2 and the mass number is decreases by _____.
Answer:
decreases, 4.

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Question 17.
In β-decay the atomic number increases by ____ unit and mass number _____.
Answer:
One, remains the same.

Question 18.
In α radiation, the charge of each alpha particle is _____.
Answer:
+2e.

Question 19.
In γ radiation, the charge of each gamma particle is _____.
Answer:
Zero.

Question 20.
In radioactive radiation, which one is travel with the speed of light _____.
Answer:
Gamma radiation.

Question 21.
\(z^{Y^{A}} \rightarrow z_{-2} Y^{A-4}+X\); Then X is _____.
Answer:
\(_{2} \mathrm{He}^{4}\) (α decay).

Question 22.
\(z^{Y^{A}} \rightarrow_{z} Y^{{A}+X}\); Then X is _____.
Answer:
γ decay.

Question 23.
The average energy released in each fission process in about _____.
Answer:
3.2 × 10-11 J.

Question 24.
Fissionable material is a radioactive element, which undergoes fission in a sustained manner when it absorbs a _____.
Answer:
Neutron.

Question 25.
_____ isotope is used to detect the presence of block in blood vessels and also used for the effective functioning of the heart.
Answer:
Na24 – Radio sodium.

Question 26.
_____ is used to cure goitre.
Answer:
Radio Iodine – I131

Question 27.
_____ is used to diagnose anaemia and also to provide treatment for the same.
Answer:
Radio – iron (Fe59).

Question 28.
Radio cobalt (Co60) and radio gold (Au198) are used in the treatment of _____.
Answer:
Skin cancer.

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Question 29.
_____ are used to sterilize the surgical devices as they can kill the germs and microbes.
Answer:
Radiations.

Question 30.
The age of the earth, fossils, old paintings and monuments can be determined by _____. technique.
Answer:
Radiocarbon dating.

Question 31.
When the body is exposed to about 600 R, it leads to _____.
Answer:
Death.

Question 32.
Radioactive materials should be kept in a thick – walled container of _____.
Answer:
Lead.

Question 33.
_____ is used to remove the heat produced in the reactor core, to produce steam.
Answer:
Coolant.

Question 34.
The abbreviation of BARC is _____.
Answer:
Bhabha Atomic Research Centre.

Question 35.
India’s 1st nuclear power station is _____.
Answer:
Tarapur Atomic Power Station.

Question 36.
The first nuclear reactor built in India was _____.
Answer:
Apsara.

Question 37.
The total nuclear power operating sites in India is _____.
Answer:
7

Question 38.
The energy released in a nuclear fission process is about ______
Answer:
200 Mev.

Question 39.
The number of \({ { _{ 0 }{ n } } }^{ 1 }\) released on an average per fission is _____.
Answer:
2.5.

Question 40.
A hydrogen bomb is based on the principle of _____.
Answer:
Nuclear fusion.

III. Match the following

Question 1.

1. Natural radioactivity (a) 3.7 × 1010 decay/second
2. Artificial radioactivity (b) spontaneous process
3. 1 curie (c) 106 decay/second
4. 1 Rd (Rutherford) (d) induced process

Answer:
1. (b) spontaneous process
2. (d) induced process
3. (a) 3.7 × 1010 decay / second
4. (c) 106 decay / second

Question 2.

1. Charge of each α particle (a) γ ray
2. Charge of each β particle (b) +2e
3. Penetration power is maximum (c) α ray
4. Ionisation power is maximum (d) zero

Answer:
1. (b) +2e
2. (d) zero
3. (a) γ ray
4. (e) α ray

Question 3.

1. Deuterium (a) \(-1^{e^{0}}\)
2. Protium (b) \(_{1} \mathrm{H}^{3}\)
3. Tritium (c) \(_{2} \mathrm{H}^{4}\)
4. α – decay (d) \(_{1} \mathrm{H}^{1}\)
5. β – decay (e) \(_{1} \mathrm{H}^{2}\)

Answer:
1. (e) \(_{1} \mathrm{H}^{2}\)
2. (d) \(_{1} \mathrm{H}^{1}\)
3. (b) \(_{1} \mathrm{H}^{3}\)
4. (c) \(_{2} \mathrm{H}^{4}\)
5. (a) \(-1^{e^{0}}\)

Question 4.

1. Uranium core bomb (a) fusion bomb
2. Plutonium core bomb (b) fission bomb
3. Hydrogen bomb (c) Nagasaki
4. Atom bomb (d) Hiroshima

Answer:
1. (d) Hiroshima
2. (c) Nagasaki
3. (a) fusion bomb
4. (b) fission bomb

Question 5.

1. Radio iron (Fe59) (a) treatment of skin diseases
2. Radio phosphorous (P32) (b) smoke detector
3. Radio gold (Au198) (c) diagnose anaemia
4. An isotope of Americium (Am241) (d) treatment of skin cancer

Answer:
1. (c) diagnose anaemia
2. (a) treatment of skin diseases
3. (d) treatment of skin cancer
4. (b) smoke detector

IV. Arrange the following in the correct sequence

Question 1.
Arrange α, β, γ rays in ascending order, on the basis of their penetrating power?
Answer:
Ascending order:

  • Alpha (α)
  • Beta (β)
  • Gamma (γ)

Question 2.
Arrange in ascending and descending order, on the basis of their Ionisation power.
Alpha (α), Beta (β), Gamma (γ)
Answer:

  1. Ascending order: Gamma (γ), Beta (β), Alpha (α)
  2. Descending order: Alpha (α), Beta (β), Gamma (γ)

Question 3.
Arrange in ascending and descending order, on the basis of their biological effect.
Alpha (α), Gamma (γ), Beta (β)
Answer:

  1. Ascending order: Alpha (α), Beta (β), Gamma (γ)
  2. Descending order: Gamma (γ), Beta (β), Alpha (α).

V. Numerical Problems

Question 1.
\(_{92} U^{238}\) emits 8α particles and 6β particles. What is the neutron / proton ratio in the product nucleus?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 7

Question 2.
The element with atomic number 84 and mass number 218 change to another element with atomic number 84 and mass number 214. The number of α and β particles emitted are respectively?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 8
Number of alpha decay, x = 1
Number of beta decay, y = 2.

Question 3.
The number of α and β particles emitted in the nuclear reaction \(_{90} \mathrm{Th}^{228} \longrightarrow_{83} \mathrm{Bi}^{12}\) are respectively.
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 9
Number of α decay, x = 4
Number of β decay, y = 1.

VI. Assertion and Reason Type Questions

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If Assertion is true, but the reason is false.
(d) If Assertion is false, but the reason is true.
(e) If the Assertion and reason both are false.

Question 1.
Assertion: All the radioactive element are ultimately converted in lead.
Reason: All the elements above lead are unstable.
Answer:
(c) If Assertion is true, but the reason is false.
Explanation: When they are converted into a lead, the emission is stopped because the nucleus of lead is stable (or lead is most stable elements in radioactive series)

Question 2.
Assertion: Among the alpha, beta and gamma-ray a particle has maximum penetrating power.
Reason: The alpha particle is heavier than beta and gamma rays.
Answer:
(e) If the Assertion and reason both are false.
Explanation: The penetrating power is maximum in case of gamma rays because gamma rays are electromagnetic radiation of very small wavelength.

Question 3.
Assertion: The ionising power of β – particle is less compared to α – particles but their penetrating power is more.
Reason: The mass of β-particle is less than the mass of α-particle
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: β – particle being emitted with very high speed compared to α – particle. Due to this reason, their loss of energy is very slow and they can penetrate the medium through a sufficient depth.

Question 4.
Assertion: Neutrons penetrate matter more readily as compared to protons.
Reason: Neutrons are slightly more massive than protons.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Neutron is about 0.1 % more massive than a proton. But the unique thing about the neutron is that while it is heavy, it has no charge (it is neutral). This lack of charge gives it the ability to penetrate matter without interacting as quickly as the beta particles or alpha particles.

Question 5.
Assertion: \(_{z} X^{A}\) undergoes a decays and the daughter product is \({ _{ z-2 } }Y^{ A-4 }\)
Reason: In α – decay, the mass number decreases by 4 and atomic number decreases by.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: \(_{z} \mathrm{X}^{\mathrm{A}} \longrightarrow_{z-2} \mathrm{X}^{\mathrm{A}-4}+_{2} \mathrm{He}^{4}\) (α decay)

Question 6.
Assertion: Moderator is used to slowing down the high energy neutrons to provide slow neutrons.
Reason: Cadmium rods are used as control rods.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Graphites and heavy water are commonly used moderators. This helps in moderator to slow down the fast neutrons.

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Question 7.
Assertion: Alpha, beta and gamma radiations are emitted.
Reason: Nuclear fission process can be performed at room temperature.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: At room temperature, the nuclear fission process can perform breaking up of heavier nucleus into two smaller nuclei. In this process to emitted the alpha, beta and gamma radiations.

Question 8.
Assertion: An enormous amount of energy is released which is called stellar energy.
Reason: Fusion reaction that takes place in the cores of the Sun and other stars.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy.

Question 9.
Assertion: Artificial radioactivity is a controlled process.
Reason: It is a spontaneous process – natural radioactivity.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Artificial radioactivity is a controlled process. It is an induced process and man-made radioactivity.

Question 10.
Assertion: Gamma rays, penetrates through materials most effectively.
Reason: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation. This is a reason they can penetrate through materials most effectively.

VII. Answer the following questions

Question 1.
Define ‘Radioactivity’.
Answer:
The phenomenon of nuclear decay of certain elements with the emission of radiations like alpha, beta, and gamma rays is called ‘radioactivity’.

Question 2.
By whom radioactivity is detected in pitchblende?
Answer:
Marie curie and Purie curie.

Question 3.
Define ‘Artificial Radioactivity’.
Answer:
The phenomenon by which even light elements are made radioactive, by artificial or induced methods, is called ‘Artificial radioactivity’ or ‘Man – made radioactivity’.

Question 4.
Define ‘One curie’.
Answer:
It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of 1 g of radium 226.
Curie = 3.7 × 1010 disintegrations per second.

Question 5.
In which elements artifical radioactivity is induced?
Answer:
Boron and aluminum

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Question 6.
What is alpha decay (α decay)? give an example.
Answer:
A nuclear reaction in which an unstable parent nucleus emits an alpha particle and forms a stable daughter nucleus is called ‘alpha decay’.
E.g. Decay of uranium (U238) to thorium (Th234) with the emission of an alpha particle.
\(_{92} \mathrm{U}^{238} \rightarrow_{90} \mathrm{Th}^{234}+_{2} \mathrm{He}^{4}\) (α – decay).

Question 7.
What is beta decay (β decay)? Give an example?
Answer:
A nuclear reaction, in which an unstable parent nucleus emits a beta particle and forms a stable daughter nucleus, is called ‘beta decay’.
E.g. Beta decay of phosphorous.
\(_{15} \mathrm{P}^{32} \rightarrow_{16} \mathrm{S}^{32}+_{-1} \mathrm{e}^{0}\) (β – decay)

Question 8.
What is gamma decay (γ decay)?
Answer:
In a γ – decay, only the energy level of the nucleus changes. The atomic number and mass number of the radioactive nucleus remain the same.

Question 9.
State the value of Roentgen in terms of Coulomb.
Answer:
Roentgen = 2.58 × 10-4 Coulomb in / kg of air.

Question 10.
Define ‘nuclear fission’ Give an example.
Answer:
The process of breaking (splitting) up of a heavier nucleus into two smaller nuclei with the release of a large amount of energy and a few neutrons are called ‘nuclear fission’.
E.g. Nuclear fission of a uranium nucleus (U235)
\(92^{\mathrm{U}^{235}}+_{0} \mathrm{n}^{1} \rightarrow_{56} \mathrm{Ba}^{141}+_{36} \mathrm{Kr}^{92}+_{30} \mathrm{n}^{1}+\mathrm{Q}(\text { energy })\)

Question 11.
Define ‘Nuclear fusion’ Give an example.
Answer:
The process in which two light nuclei combine to form a heavier nucleus is termed as ‘Nuclear fusion’.
E.g. \(_{1} \mathrm{H}^{2}+_{1} \mathrm{H}^{2} \rightarrow_{2} \mathrm{He}^{4}+\mathrm{Q}(\text { Energy })\)

Question 12.
Write down the types of the nuclear reactor.
Answer:
Breeder reactor, fast breeder reactor, pressurized water reactor, pressurized heavy water reactor, boiling water reactor, water – cooled reactor, gas – cooled reactor, fusion reactor and thermal reactor are some types of nuclear reactors, which are used in different places worldwide.

Question 13.
What is the safe limit of receiving radioactive radiations?
Answer:
100 m R per week

VIII. Answer in the details:

Question 1.
Explain the principle and working of an atom bomb?
Answer:
Atom bomb:
(i) The atom bomb is based on the principle of the uncontrolled chain reaction. In an uncontrolled chain reaction, the number of neutrons and the number of fission reactions multiply almost in a geometrical progression.

(ii) This releases a huge amount of energy in a very small time interval and leads to an explosion.
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 10

Structure:
(i) An atom bomb consists of a piece of fissile material whose mass is subcritical. This piece has a cylindrical void.

(ii) It has a cylindrical fissile material which can fit into this void and its mass is also subcritical. When the bomb has to be exploded, this cylinder is injected into the void using a conventional explosive.

(iii) The two pieces of fissile material join to form the supercritical mass, which leads to an explosion. During this explosion, a tremendous amount of energy in the form of heat, light and radiation is released.

(iv) A region of very high temperature and pressure is formed in a fraction of a second along with the emission of hazardous radiation like y rays, which adversely affect the living creatures. This type of atom bombs was exploded in 1945 at Hiroshima and Nagasaki in Japan during World War II.

Question 2.
State and define the units of radioactivity.
Answer:
Curie : It is the traditional unit of radioactivity. It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of lg of radium 226. 1 curie = 3.7 × 1010 disintegrations per second.

Rutherford (Rd) : It is another unit of radioactivity. It is defined as the quantity of a radioactive substance, which produces 106 disintegrations in one second.
1 Rd = 106 disintegrations per second.

Becquerel (Bq) : It is the SI unit of radioactivity is becquerel. It is defined as the quantity of one disintegration per second.

Roentgen (R) : It is the radiation exposure of γ and x-rays is measured by another unit called roentgen. One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.

Question 3.
Write down the features of nuclear fission and nuclear fusion.
Answer:

Nuclear Fission Nuclear Fusion
1. The process of breaking up (splitting) of a heavy nucleus into two smaller nuclei is called ‘nuclear fission’. 1. Nuclear fusion is the combination of two lighter nuclei to form a heavier nucleus.
2. Can be performed at room temperature. 2. Extremely high temperature and pressure are needed.
3. Alpha, beta and gamma radiations are emitted. 3. Alpha rays, positrons, and neutrinos are emitted.
4. Fission leads to emission of gamma radiation. This triggers the mutation in the human gene and causes genetic transform diseases. 4. Only light and heat energy are emitted.

Question 4.
Write down the medical and industrial application of radioisotopes?
Answer:

  1. Radio sodium (Na24) is used for the effective functioning of the heart.
  2. Radio – Iodine (I131) is used to cure goitre.
  3. Radio – Iron is (Fe59) is used to diagnose anaemia and also to provide treatment for the same.
  4. Radio Phosphorous (P32) is used in the treatment of skin diseases.
  5. Radio Cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
  6. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.
  7. Radio cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
  8. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.

Question 5.
Write a note about stellar energy.
Answer:
The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy. Where does this high energy come from? All-stars contain a large amount of hydrogen. The surface temperature of the stars is very high which is sufficient to induce fusion of the hydrogen nuclei.

Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as ‘stellar energy’. Thus, nuclear fusion or thermonuclear reaction is the source of light and heat energy in the Sun and other stars.

IX. Additional HOT Questions

Question 1.
Why is neutron so effective as bombarding particle?
Answer:
A neutron carries no charge. It easily penetrates even a heavy nucleus without being repelled or attracted by nucleus and electrons. So it serves as an ideal projectile for starting a nuclear reaction.

Question 2.
Is there any difference between electron and a beta particle.
Answer:
Basically, there is no difference between an electron and a beta particle. β particle is the name given to an electron emitted from the nucleus.

Question 3.
Why are the control rods made of cadmium?
Answer:
Cadmium has high cross – section for the absorption of neutrons.

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Question 4.
Name two radioactive elements that are not found in observable quantities why is it so?
Answer:
Tritium and Plutonium are two radioactive elements that are not found in observable quantities in the universe.
It is because half-life period of each of two elements is very short compared to the age of the universe.

Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics Read More »

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

Question 1.
Find the combined equation of the straight lines whose separate equations are x – 2y – 3 = 0 and x + y + 5 = 0.
Solution:
The given separate equations of the lines are
x – 2y – 3 = 0 and x + y + 5 = 0
∴ The combined equation of the straight lines is
(x – 2y – 3 ) (x + y + 5) = 0
x2 + xy + 5x – 2xy – 2y2 – 10y – 3x – 3y – 15 = 0
x2 – xy – 2y2 + 2x – 13y – 15 = 0

Question 2.
Show that 4x2 + 4xy + y2 – 6x – 3y – 4 = 0 represents a pair of parallel lines.
Solution:
Comparing this equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
we get a = 4, h = \(\frac{4}{2}\) = 2 , b = 1, g = – 3, f = – 3/2, c = – 4
The condition for the lines to be parallel is h2 – ab = 0
Now h2 – ab = 22 – (4) (1) = 4 – 4 = 0
h2 – ab = 0 ⇒ The given equation represents a pair of parallel lines.

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

Question 3.
Show that 2x2 + 3xy – 2y2 + 3x + y + 1 = 0 represents a pair of perpendicular lines.
Solution:
The equation of the given pair of straight lines is
2x2 + 3xy – 2y2 + 3x + y + 1 = 0 ……….. (1)
Compare this equation with the equation
ax2 + 2hxy + by2 + 2gx + 2f y + c = 0 ……… (2)
a = 2, 2h = 3, b = – 2 , 2g = 3, 2f = 1, c = 1
The condition for pair of straight lines to be perpendicular is a + b = 0.
2 – 2 = 0
Hence the given pair of lines represents a perpendicular straight lines.

Question 4.
Show that the equation 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 represents a pair of intersecting lines. Show further that the angle between them is tan-1(5).
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 50
The given equation represents a pair of straight lines.
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 59

Question 5.
Prove that the equation to the straight lines through the origin, each of which makes an angle α with the straight line y = x is x2 – 2xy sec 2α + y2 = 0
Solution:
Slope of y = x is m = tan θ = 1
⇒ θ = 45°
The new lines slopes will be
m = tan(45 + α) and m = tan (45 – α)
∴ The equations of the lines passing through the origin is given by
y = tan(45 + α)x and y = tan(45 – α)x
(i.e) y = tan(45 + α)x = 0 and y = tan(45 – α)x = 0
The combined equation is [y – tan (45 + α)x] [y – tan (45 – α)x] = 0
y2 + tan(45 + α)tan(45 – α)x2 – xy[tan(45 – α) + tan(45 + α)] = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 52
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 53
Let the equation of lines passes through the origin
So the equations are y = m1x = 0 and y = m2x = 0
So the combined equations is (y – m1x) (y – m2x) = 0
(i.e)y2 – xy(m1 + m2) + m1m2x = 0
(i.e) y2 – xy(2sec α) + x2(1) = 0
(i.e) y2 – 2xy sec 2α + x2 = 0

Question 6.
Find the equation of the pair of straight lines passing through the point (1, 3) and perpendicular to the lines 2x – 3y + 1 = 0 and 5x + y – 3 = 0
Solution:
The equation of the given lines are
2x – 3y + 1 = 0 ……….. (1)
5x + y – 3 = 0 ……….. (2)
Equation of any line perpendicular to 2x – 3y + 1 = 0 is
– 3x – 2y + k = 0
3x + 2y – k = 0
This line passes through the point (1, 3)
∴ 3(1) + 2(3) – k = 0
3 + 6 – k = 0 ⇒ k= 9
Substituting the value of k in the above equation we have
3x + 2y – 9 = 0 ………. (3)
Equation of any line perpendicular to 5x + y – 3 = 0 is
x – 5y + k1 = 0
This line passes througi the point (1 , 3)
∴ 1 – 5 (3) + k1 = 0
1 – 15 + k1 ⇒ k1 = 14
Substituting the value of k1 in the above equation we have
x – 5y + 14 = 0 ……….. (4)
The combined equation of (3) and (4) is
( 3x + 2y – 9) (x – 5y + 14 ) = 0
3x2 – 15xy + 42x + 2xy – 10y2 + 28y – 9x + 45 y – 126 = 0
3x2 – 13xy – 10y2 + 33x + 73y – 126 = 0

Question 7.
Find the separate equation of the following pair of straight lines
(i) 3x2 + 2xy – y2 = 0
(ii) 6 (x – 1)2 + 5(x – 1)(y – 2) – 4(y – 2)2 = 0
(iii) 2x2 – xy – 3y2 – 6x + 19y – 20 = 0
Solution:
(i) 3x2 + 2xy – y2 = 0
The given equation is
3x2 + 2xy – y2 = 0 ……. (1)
3x2 + 3xy – xy – y2 = 0
3x (x + y) – y (x + y) = 0
(3x – y) (x + y) = 0
3x – y = 0 and x + y = 0
∴ The separate equations are
3x – y = 0 and x + y = 0

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

(ii) 6 (x – 1)2 + 5 (x – 1)(y – 2) – 4(y – 2)2 = 0
⇒ 6(x2 – 2x +1) + 5(xy – 2x – y + 2) – 4( y2 – 4y + 4) = 0
(i.e) 6x2 – 12x + 6 + 5xy – 10x – 5y + 10 – 4y2 + 16y – 16 = 0
(i.e) 6x2 + 5xy – 4y2 – 22x + 11y = 0
Factorising 6x2 + 5xy – 4y2 we get
6x2 – 3xy + 8xy – 4y2 = 3x (2x – y) + 4y (2x – y)
= (3x + 4y)(2x – y)
So, 6x2 + 5xy – 4y2 – 22x + 11y = (3x + 4y + l )(2x – y + m)
Equating coefficient of x ⇒ 3m + 21 = -22 …….. (1)
Equating coefficient of y ⇒ 4m – l = 11 ……. (2)
Solving (1) and (2) we get l = -11, m = 0
So the separate equations are 3x + 4y – 11 = 0 and 2x – y = 0

(iii) 2x2 – xy – 3y2 – 6x + 19y – 20 = 0
Factorising 2x2 – xy – 3y2 we get
2x2 – xy – 3y2 = 2x2 + 2xy – 3xy – 3y2
= 2x(x + y) – 3y(x + y) = (2x – 3y) (x + y)
∴ 2x2 – xy – 3y2 – 6x + 19y – 20 = (2x – 3y + l)(x + y + m)
Equating coefficient of x 2m + l = -6 ……. (1)
Equating coefficient of y -3m + l = 19 …….. (2)
Constant term -20 = lm
Solving (1) and (2) we get l = 4 and m = – 5 where lm = – 20.
So the separate equations are 2x – 3y + 4 = 0 and x + y – 5 = 0

Question 8.
The slope of one of the straight lines ax2 + 2hxy + by2 = 0 is twice that of the other, show that 8h2 = 9ab.
Solution:
ax2 + 2hxy + by2 = 0
We are given that one slope is twice that of the other.
So let the slopes be m and 2m.
Now sum of the slopes = m + 2m
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 40

Question 9.
The slope of one of the straight lines ax2 + 2hxy + by2 = 0 is three times the other, show that 3h2 = 4ab.
Solution:
Let the slopes be m and 3m.
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 41

Question 10.
A ∆OPQ is formed by the pair of straight lines x2 – 4xy + y2 = 0 and the line PQ. The equation of PQ is x + y – 2 = 0. Find the equation of the median of the triangle ∆OPQ drawn from the origin O.
Solution:
Equation of pair of straight lines is x2 – 4xy + y2 = 0 ….. (1)
Equation of the given line is x + y – 2 = 0 ⇒ y = 2 – x ……… (2)
On solving (1) and (2) we get x2 – 4x (2 – x) + (2 – x)2 = 0
(i.e) x2 – 8x + 4x2 + 4 + x2 – 4x = 0
(i.e) 6x2 – 12x + 4 = 0
(÷ by 2) 3x2 – 6x + 2 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 54
The midpoint of PQ is
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 43

Question 11.
Find p and q, ¡f the following equation represents a pair of perpendicular lines 6x2 + 5xy – py2 + 7x + qy – 50
Solution:
6x2 + 5xy – py2 + 7x + qy – 50
The given equation represents a pair of perpendicular lines
⇒ coefficient of x2 + coefficient of y2 = 0
(i.e) 6 – p = 0 ⇒ p = 6
Now comparing the given equation with the general form
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
we get a = 6, b = -6 and c = -5, f = q/2, g = 7/2 and h = 5/2
The condition for the general form to represent a pair of straight lines is abc + 2fgh – af2 – bg2 – ch2 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 44

Question 12.
Find the value of k, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting, 12x2 + 7xy – 12y2 – x + 7y + k = 0.
Solution:
Comparing the given equation with the general form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
we get a = 12, b = -12, c = k, f = 7/2, g = – 1/2, h = 7/2
Here a + b = 0 ⇒ the given equation represents a pair of perpendicular lines
To find k: The condition for the given equation to represent a pair of straight lines is abc + 2fgh – af2 – bg2 – ch2 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 46
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 47

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

Question 13.
For what value of k does the equation 12x2 + 2kxy + 2y2, + 11x – 5y + 2 = 0 represent two straight lines.
Solution:
12x2 + 2 kxy + 2y2 + 11x – 5y + 2 = 0
Comparing this equation with the general form we get
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 48
4k2 + 55k + 175 = 0
4k2 + 20k + 35k + 175 = 0
4k(k + 5) + 35(k + 5) = 0
(4k + 35) (k + 5) = 0
k = -5 or -35/4

Question 14.
Show that the equation 9x2 – 24xy + 16y2 – 12x + 16y – 12 = 0 represents a pair of parallel lines. Find the distance between them.
Solution:
Comparing the given equation with ax2 + 2kxy + by2 = 0 we get a = 9, h = -12, b = 16.
Now h2 = (-12)2 = 144, ab = (9) (16) = 144
h2 = ab ⇒ The given equation represents a pair of parallel lines.
To find their separate equations:
9x2 – 24xy + 16y2 = (3x – 4y)2
So, 9x2 – 24xy +16y2 – 12x + 16y – 12 = (3x – 4y + l )(3x – 4y + m)
Here coefficient of x ⇒ 3m + 3l = -12 ⇒ m + l = -4
coefficient of y ⇒ -4m – 4l = 16 ⇒ m + l = -4
Constant term l m = -12
Now l + m = -4 and lm = -12 ⇒ l = -6 and m = 2
So the separate equations are 3x – 4y – 6 = 0 and 3x – 4y + 2 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 49

Question 15.
Show that the equation 4x2 + 4xy + y2 – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them.
Solution:
4x2 + 4xy + y2 – 6x – 3y – 4 = 0
a = 4,
b = 1,
h = 4/2 = 2
h2 – ab = 22 – (4) (1) = 4 – 4 = 0
⇒ The given equation represents a pair of parallel lines.
To find the separate equations 4x2 + 4xy + y2 = (2x + y)2
So, 4x2 + 4xy + y2 – 6x – 3y – 4 = (2x + y + l )(2x + y + m)
Coefficient of x ⇒ 2m + 2l = -6 ⇒ l + m = – 3 ……. (1)
Coefficient of y ⇒ l + m = – 3 ……… (2)
Constant term ⇒ l m = – 4 ……… (3)
Now l + m = -3 and lm = -4 ⇒ l = -4, m = 1
So the separate equations are 2x + y + 1 = 0 and 2x + y – 4 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 55

Question 16.
Prove that one of the straight lines given by ax2 + 2hxy + by2 = 0 will bisect the angle between the co-ordinate axes if (a + b)2 = 4h2.
Solution:
Let the slopes be l and m
∵ One line bisects the angle between the coordinate axes ⇒ θ = 45°
So tan θ = 1
The slopes are l and m
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 56

Question 17.
If the pair of straight lines x2 – 2kxy – y2 = 0 bisect the angle between the pair of straight lines x2 – 2lxy – y2 = 0, show that the latter pair also bisects the angle between the former.
Solution:
Given that x2 – 2kxy – y2 = 0 …….. (1)
Bisect the angle between the lines x2 – 11xy – y2 = 0 …… (2)
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 57
x2 – 2kxy – y2 = 0

Question 18.
Prove that the straight lines joining the origin to the points of intersection of 3x2 + 5xy – 3y2 + 2x + 3y = 0 and 3x – 2y – 1 = 0 are at right angles.
Solution:
The equation of the pair of straight lines is
3x2 + 5xy – 3y2 + 2x + 3y = 0 ……… (1)
The given line is 3x – 2y – 1 = 0
3x – 2y = 1 ……….. (2)
The equation of the straight lines joining the origin to the points of intersection of the pair of lines (1) and the line (2) is obtained by homogeneous using equation (1) by using equation (2)
(1) ⇒ (3x2 + 5xy – 3y2) + (2x + 3y)(1) = 0
(3x2 + 5xy – 3y2) + (2x + 3y) (3x – y) = 0
3x2 + 5xy – 3y2 + 6x2 – 4xy + 9xy – 6y2 = 0
9x2 + 10xy – 9y2 = 0 ………. (3)
Coefficient of x2 + coefficient of y2 = 9 – 9 = 0
∴ The pair of straight line (3) represents a perpendicular straight lines.

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Additional Questions Solved

Question 1.
Find the angle between the pair of straight lines given by (a2 – 3b2)x2 + 8abxy + (b2 – 3a2)y2 = 0.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 58

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

Question 2.
Show that 9x2 + 24xy + 16y2 + 21x + 28y + 6 = 0 represents a pair of parallel straight lines and find the distance between them.
Solution:
9x2 + 24xy + 16y2 + 21x + 28y + 6 = 0
Here a = 9.6,
b = 16,
g = \(\frac{21}{2}\),
f = 14,
c = 6,
h = 12
h2 – ab = (12)2 – 9(16) = 144 – 144 = 0
∴ The lines are parallel.
9x2 + 24xy + 16y2 = (3x + 4y)(3x + 4y)
Let 9x2 + 24xy + 16y2 + 21x + 28y + 6 = (3x + 4y + l)(3x + 4y + m)
Equating the coefficients of x and constant term
3l + 3m = 21
lm = 6
Solving we get, l = 1 or 6
m = 6 or 1
∴ The separate equations are 3x + 4y + 1 = 0 and 3x + 4y + 6 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 7

Question 3.
If the equation 12x2 – 10xy + 2y2 + 14x – 5y + c = 0 represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle
between them.
Solution:
12x2 – 10xy – 2y2 + 14x – 5y + c = 0
ax2 + 2hxy + by2 +2gx + 2fy – c = 0
Here a = 12,
b = 2,
g = 7,
f = 5/2,
c = c,
h = -5
af2 + bg2 + ch2 – 2fgh – abc = 0 is the condition
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 8
The equation is 12x2 – 10y + 2y2 + 14x – 5y + 2 = 0
12x2 – 10xy + 2y = (3x – y)(4x – 2y)
Let 12x2 – 10y + 2y2 + 14x – 5y + 2(3x – y + l)(4x – 2y + m)
So that 4l + 3m = 14 , -2l – m = -5
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 9

Question 4.
For what value of k does 12x2 + 7xy + ky2 + 13x – y + 3 = 0 represents a pair of straight lines? Also write the separate equations.
Solution:
12x2 + 7xy + ky2 + 13x – y + 3 = 0
a = 12,
h = \(\frac{7}{2}\),
f = \(-\frac{1}{2}\) ,
c = 3
af2 + bg2 + ch2 – abc – 2fgh = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 11
⇒ 12 + 169k + 147 – 144k + 91 = 0
25k = – 250 ⇒ k = -10
The equation is 12x2 + 7xy – 10y2 + 13x – y + 3 = 0
To find separate equations: 12x2 + 7xy – 10y2 = (3x – 2y)(4x + 5y)
Let 12x2 + 7xy – 10y2 + 13x – y + 3 = 0(3x – 2y + l)(4x + 5y + m)
Equating the coefficient of x ⇒ 4l + 3m = 13 …… (1)
Equating the coefficient of y ⇒ 5l – 2m = -1 …… (2)
(1) × 2 ⇒ 8l + 6m = 26
(2) × 3 ⇒ 15l – 6m = -3
23l = 23 ⇒ l = 1
4 + 3 m = 13
3 m = 9 ⇒ m = 3
The separate equations are 3x – 2y + 1 = 0 and 4x + 5y + 3 = 0

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4

Question 5.
Show that 3x2 + 10xy + 8y2 + 14x + 22y + 15 = 0 represents a pair of straight lines and the angle between them is tan-1\(\left(\frac{2}{11}\right)\)
Solution:
3x2 + 10xy + 8y2 + 14x + 22y + 15 = 0
a = 3,
A = 5,
b = 8,
g = 7,
f = 11,
c = 15
The condition is af2 + bg2 + ch2 – abc – 2fgh = 0
3(11)2 + 8(7)2 + 15 (5)2 – (3)(8)(15) – 2(11)(7)(5) = 363 + 392 + 375 – 360 – 770 = 0
The angle between the pair of straight line is given by
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 12

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Read More »

Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions

Samacheer Kalvi 10th Science Types of Chemical Reactions Textual Problems Solved

Question 1.
Calculate the pH of 0.01 M HNO3?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 1

Question 2.
The hydroxyl ion concentration of a solution is 1 × 10-9 M. What is the pOH of the solution?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 2

Question 3.
A solution has a pOH of 11.76. What is the pH of this solution?
Solution:
pH = 14 – pOH
pH = 14 – 11.76 = 2.24.

Question 4.
Calculate the pH of 0.001 molar solution of HCl.
Solution:
HCl is a strong acid and is completely dissociated in its solutions according to the process:
\(\mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
From this process it is clear that one mole of HCl would give one mole of H+ ions.
Therefore, the concentration of H+ ions would be equal to that of HCl, i.e., 0.001 molar or 1.0 × 10-3 mol litre-1.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 3

Question 5.
What would be the pH of an aqueous solution of sulphuric acid which is 5 × 10-5 mol litre-1 in concentration?
Solution:
Sulphuric acid dissociates in water as:
\(\mathrm{H}_{2} \mathrm{SO}_{4(\mathrm{aq})} \rightarrow 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)
Each mole of sulphuric acid gives two moles of H+ ions in the solution. One litre of H2SO4 solution contains 5 × 10-5 moles of H2SO4 which would give 2 × 5 × 10-5 = 10 × 10-5 or 1.0 × 10-4 moles of H+ ion in one litre of the solution.
Therefore,
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 36

Question 6.
Calculate the pH of 1 × 10-4 molar solution of NaOH.
Solution:
NaOH is a strong base and dissociates in its solution as:
\(\mathrm{NaOH}_{(\mathrm{aq})} \rightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
One mole of NaOH would give one mole of OH ions.
Therefore,
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 4

Question 7.
Calculate the pH of a solution in which the concentration of the hydrogen ions is 1.0 × 10-8 mol litre-1.
Solution:
Here, although the solution is extremely dilute, the concentration given is not of an acid or a base but that of H+ ions. Hence, the pH can be calculated from the relation:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 5

Question 8.
If the pH of a solution is 4.5, what is its pOH?
Solution:
pH + pOH = 14
⇒ pOH = 14 – 4.5 = 9.5
⇒ pOH = 9.5.

Samacheer Kalvi 10th Science Types of Chemical Reactions Textual Evaluation Solved

I. Choose the correct answer.

Question 1.
\(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})}\) is a ______.
(a) Decomposition Reaction
(b) Combination Reaction
(c) Single Displacement Reaction
(d) Double Displacement Reaction.
Answer:
(b) Combination Reaction
Hint: It is a combination reaction between H2 and Cl2 to form HCl as product.

Question 2.
Photolysis is a decomposition reaction caused by:
(a) heat
(b) electricity
(c) light
(d) mechanical energy
Answer:
(c) light

Question 3.
A reaction between carbon and oxygen is represented by \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}\) + Heat. In which of the type(s), the above reaction can be classified?
(i) Combination Reaction
(ii) Combustion Reaction
(iii) Decomposition Reaction
(iv) Irreversible Reaction
(a) i and ii
(b) i and iv
(c) i, ii and iii
(d) i, ii and iv.
Answer:
(d) i, ii and iv.
Hint:

  • It is a combination reaction with one product.
  • It involves the combustion of carbon to form CO2.
  • The reaction is irreversible and we cannot reverse the reaction.

SamacheerKalvi.Guru

Question 4.
The chemical equation
Na2SO4(aq) + BaCI2(aq) → BaSO4(s)↓ + 2NaCl(aq) represents which of the following types of reaction?
(a) Neutralisation
(b) Combustion
(c) Precipitation
(d) Single displacement
Answer:
(c) Precipitation

Question 5.
Which of the following statements are correct about a chemical equilibrium?
(i) It is dynamic in nature
(ii) The rate of the forward and backward reactions are equal at equilibrium
(iii) Irreversible reactions do not attain chemical equilibrium
(iv) The concentration of reactants and products may be different
(a) i, ii and iii
(b) i, ii and iv
(c) ii, iii and iv
(d) i, iii and iv.
Answer:
(a) i, ii and iii
Hint: Chemical equilibrium is dynamic in nature for a reversible reaction.
At equilibrium, Rate of forwarding reaction = Rate of backward reaction.

Question 6.
A single displacement reaction is represented by
X(s) + 2HCl(aq) → XCl2(aq) + H2(g). Which of the following(s) could be X?
(i) Zn
(ii) Ag
(iii) Cu
(iv) Mg. Choose the best pair.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)

Question 7.
Which of the following is not an “element + element → compound” type reaction?
(a) \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}\)
(b) \(2 \mathrm{K}_{(\mathrm{s})}+\mathrm{Br}_{2(l)} \rightarrow 2 \mathrm{KBr}_{(\mathrm{s})}\)
(c) \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{g})}\)
(d) \(4 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3(\mathrm{s})}\).
Answer:
(c) \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{g})}\)
Hint: It involves one reactant compound i.e., carbon monoxide (CO).
\(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2}\)

Question 8.
Which of the following represents a precipitation reaction?
(a) \(\mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{s})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{s})}\)
(b) \(\mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{aq})}+\mathrm{D}_{(\mathrm{l})}\)
(c) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{aq})}\)
(d) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{s})} \rightarrow \mathrm{C}_{(\mathrm{aq})}+\mathrm{D}_{(\mathrm{l})}\).
Answer:
(c) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{aq})}\)
Hint: It involves the formation of solid precipitation by mixing of two aqueous solutions.

Question 9.
The pH of a solution is 3. Its [OH] concentration is:
(a) 1 × 10-3 M
(b) 3 M
(c) 1 × 10-11 M
(d) 11 M
Answer:
(c) 1 × 10-11 M

Question 10.
Powdered CaCO3 reacts more rapidly than flaky CaCO3 because of ______.
(a) large surface area
(b) high pressure
(c) high concentration
(d) high temperature.
Answer:
(a) large surface area
Hint: We know that greater the surface area, faster will be a chemical reaction. Hence powdered CaCO3 reacts more rapidly.

II. Fill in the blanks.

Question 1.
A reaction between an acid and a base is called ______.
Answer:
Neutralization.

Question 2.
When lithium metal is placed in hydrochloric acid, _______ gas is evolved.
Answer:
Hydrogen.

Question 3.
The equilibrium attained during the melting of ice is known as ______.
Answer:
Physical equilibrium.

SamacheerKalvi.Guru

Question 4.
The pH of a fruit juice is 5.6. If you add slaked lime to this juice, its pH ______ (increase/decrease)
Answer:
Increases.

Question 5.
The value of the ionic product of water at 25°C is ______.
Answer:
1 × 10-14

Question 6.
The normal pH of human blood is ______.
Answer:
7.4.

Question 7.
Electrolysis is type of _______ reaction.
Answer:
Decomposition reaction.

Question 8.
The number of products formed in a synthesis reaction is ______.
Answer:
One.

Question 9.
Chemical volcano is an example for ______ type of reaction.
Answer:
Decomposition reaction.

Question 10.
The ion formed by dissolution of H+ in water is called ______.
Answer:
Hydronium ion.

III. Match the following

Question 1.
Identify the types of reaction.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 6
Answer:
i – c, ii-a, iii – d, iv- b.

IV. True or False: (If false give the correct statement)

Question 1.
Silver metal can displace hydrogen gas from nitric acid.
Answer:
False.
Correct statement: In the activity series, any metals that are below hydrogen will not react with HNO3.

Question 2.
The pH of rainwater containing dissolved gases like SO3, CO2, NO2 will be less than 7.
Answer:
True.

Question 3.
At the equilibrium of a reversible reaction, the concentration of the reactants and the products will be equal.
Answer:
False.
Correct statement: At equilibrium rate of the forward reaction is equal to the rate of backward reaction.

Question 4.
Periodical removal of one of the products of a reversible reaction increases the yield.
Answer:
True.

Question 5.
On dipping a pH paper in a solution, it turns into yellow. Then the solution is basic.
Answer:
True.

V. Short Answer Questions

Question 1.
When an aqueous solution of potassium chloride is added to an aqueous solution of silver nitrate, a white precipitate is formed. Give the chemical equation of this reaction.
Answer:
\(\mathrm{KCl}_{(\mathrm{aq})}+\mathrm{AgNO}_{3(\mathrm{aq})} \rightarrow \mathrm{AgCl}_{(\mathrm{s})}+\mathrm{KNO}_{3(\mathrm{aq})}\)
Formation of white precipitate by the above reaction is due to formation of silver chloride (AgCl).

Question 2.
Why does the reaction rate of a reaction increase on raising the temperature?
Answer:
The rate of a reaction increases at higher temperature, because the heat to the reactants provide energy to breakup more bonds and speeds up the reaction.

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Question 3.
Define a combination reaction. Give one example of an exothermic combination reaction.
Answer:
A chemical reaction in which 2 or more reactants combine to form a single product, the reaction is known as combination reaction. Most of the combination reaction are exothermic because they involve formation of new bonds. For example,
\(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})}\)
\(\mathrm{SiO}_{2(\mathrm{s})}+\mathrm{CaO}_{(\mathrm{s})} \rightarrow \mathrm{CaSiO}_{3(\mathrm{s})}\).

Question 4.
Differentiate reversible and irreversible reactions.
Answer:

Reversible reaction Irreversible reaction
1. Reactions can be reversed 1. The reaction cannot be reversed
2. It proceeds in both directions 2. It is unidirectional
3. It attains equilibrium 3. Equilibrium is not attained
4. It is relatively slow 4. It is fast

VI. Answer in detail.

Question 1.
What is called thermolysis reactions?
Answer:
A chemical reaction is a process in which old bond breaks up and new chemical bond get formed. Thermolysis chemical reactions is a special type of chemical reaction in which the reactant get decomposed by heat. For example,
\(\mathrm{CaCO}_{3(\mathrm{s})} \stackrel{\text { Heat }}{\rightleftharpoons} \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{g})}\)
\(2 \mathrm{HgO}_{(\mathrm{s})} \stackrel{\mathrm{Heat}}{\longrightarrow} 2 \mathrm{Hg}_{(\mathrm{l})}+\mathrm{O}_{2(\mathrm{g})}\)
In these reactions heat is supplied to break the bonds, so generally they are endothermic in nature.

Question 2.
Explain the types of double displacement reactions with examples.
Answer:
When two compounds react, if their ions are interchanged, then the reaction is called double displacement reactions. There are two types of double displacement reactions. They are
(i) Precipitation reactions : When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction. Because the insoluble compound, formed as one of the products, is a precipitate and hence the reaction is so called.

(ii) When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, a double displacement reaction takes place between them.
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s)↓ + 2KNO3(aq).

(iii) Potassium and lead displace or replace one other and form a yellow precipitate of lead (II) iodide.

(iv) Neutralization reactions: When an acid reacts with the base to form a salt and water. It is called ‘neutralization reaction’ as both acid and base neutralize each other.

(v) Reaction of sodium hydroxide with hydrochloric acid is a typical neutralization reaction. Here, sodium replaces hydrogen from hydrochloric acid forming sodium chloride, a neutral soluble salt.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Question 3.
Explain the factors influencing the rate of a reaction.
Answer:
The factors influencing the rate of a reaction are,
(i) Nature of the reactants: The reaction of sodium with hydrochloric acid is faster than that with acetic acid. Do you know why? Hydrochloric acid is a stronger acid than acetic acid and thus more reactive. So, the nature of the reactants influences the reaction rate.
\(\begin{aligned} 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} & \rightarrow 2 \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}(\mathrm{fast}) \\ 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} & \rightarrow 2 \mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}(\mathrm{slow}) \end{aligned}\).

(ii) The concentration of the reactants: Changing the number of reactants also increases the reaction rate. The amount of the substance present in a certain volume of the solution is called ‘concentration’. More the concentration, more particles per volume exist in it and hence faster the reaction. Granulated zinc reacts faster with 2M hydrochloric acid than 1M hydrochloric acid.

(iii) Temperature: Most of the reactions go faster at a higher temperature. Because adding heat to the reactants provides energy to break more bonds and thus speed up the reaction. Calcium carbonate reacts slowly with hydrochloric acid at room temperature. When the reaction mixture is heated the reaction rate increases.

(iv) Pressure: If the reactants are gases, increasing their pressure increases the reaction rate. This is because on increasing the pressure the reacting particles come closer and collide frequently.

(v) Catalyst: A catalyst is a substance which increases the reaction rate without being consumed in the reaction. In certain reactions, adding a substance as catalyst speeds up the reaction. For example, on heating potassium chlorate, it decomposes into potassium chloride and oxygen gas, but at a slower rate. If manganese dioxide is added, it increases the reaction rate.

(vi) The surface area of the reactants: When solid reactants are involved in a reaction, their powdered form reacts more readily. For example, powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips. Because powdering of the reactants increases the surface area and more energy is available on the collision of the reactant particles. Thus, the reaction rate is increased. You will study more about reaction rate in your higher classes.

Question 4.
How does pH play an important role in everyday life?
Answer:
Our body works within the pH range of 7.0 to 7.8. Living organisms can survive only in a narrow range of pH change. Different body fluids have different pH values.
Eg: pH of blood is ranging from 7.35 to 7.45. Any increase or decrease in this value leads to diseases. The ideal pH for blood is 7.4.

pH in our digestive system : It is very interesting to note that our stomach produces hydrochloric acid. It helps in the digestion of food without harming the stomach. During indigestion the stomach produces too much acid and this causes pain and irritation. pH of the stomach fluid is approximately 2.0.

pH changes as the cause of tooth decay : pH of the saliva normally ranges between 6.5 to 7.5. White enamel coating of our teeth is calcium phosphate, the hardest substance in our body. When the pH of the mouth saliva falls below 5.5, the enamel gets weathered. Toothpastes, which are generally basic ate used for cleaning the teeth that can neutralise the excess acid and prevent tooth decay.

pH of soil : In agriculture, the pH of the soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.

pH of rain water : The pH of rain water is approximately 7, which means that it is neutral and also represents its high purity. If the atmospheric air is polluted with oxide gases of sulphur and nitrogen, they get dissolved in the rain water and make its pH less than 7. Thus, if the pH of rain water is less than 7, then it is called acid rain. When acid rain flows into the rivers it lowers the pH of the river water also. The survival of aquatic life in such rivers becomes difficult.

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Question 5.
What is chemical equilibrium? What are its characteristics?
Answer:
Chemical equilibrium is the state for a reversible chemical reaction where the rate of forwarding direction is equally balanced by the rate of backward direction and the process seems like to be stopped.
Its characteristics are,

  • The rate of the forward and backward reaction are equal in chemical equilibrium.
  • The observable properties such as pressure, concentration, colour, density, viscosity, etc. of the system unchanged with time.
  • In physical equilibrium, the volume of all phases remains constant.

VII. HOT Questions

Question 1.
A solid compound ‘A’ decomposes on heating into ‘B’ and a gas ‘C’ On passing the gas ‘C’ through water. It becomes acidic. Identify A, B and C.
\(\mathrm{CaCO}_{3(\mathrm{s})} \longrightarrow \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{g})}\)
Calcium carbonate decomposes into solid Calcium oxide and a gas CO2. This CO2 dissolves in water and forms carbonic acid.
\(\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}\) (Carbonic acid).

Question 2.
Can a nickel spatula be used to stir copper sulphate solution? Justify your answer.
Answer:
No, a nickel spatula cannot be used to stir CuSO4 solution, because Nickel will displace copper from CuSO3 solution and Cu gets deposited on the Ni spatula.

VIII. Solve the following problems.

Question 1.
Lemon juice has a pH = 2, what is the concentration of H+ ions?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 7

Question 2.
Calculate the pH of 1.0 × 10-4 molar solution of HNO3?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 8

Question 3.
What is the pH of 1.0 × 10-5 molar solution of KOH?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 9

Question 4.
The hydroxide ion concentration of a solution is 1 × 10-11. What is the pH of the solution?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 10

Samacheer Kalvi 10th Science Types of Chemical Reactions Additional Question Solved

I. Choose the best answer.

Question 1.
Methane + Oxygen → A + Water. Identify A ______.
(a) Carbon monoxide
(b) Carbon dioxide
(c) ethane
(d) LPG.
Answer:
(i) Carbon dioxide

Question 2.
The chemical reaction in which electricity is used to bring about the change is:
(a) Thermal decomposition
(b) Photo decomposition
(c) Single displacement reaction
(d) Electrolytic decomposition
Answer:
(d) Electrolytic decomposition

Question 3.
Decomposition of the molecule occurs on passing an electric current through its aqueous solution. This process is termed as ______.
(a) Thermolysis
(b) Photolysis
(c) Electrolysis
(d) None of these.
Answer:
(c) Electrolysis

Question 4.
The most reactive element in the activity series is:
(a) platinum
(b) potassium
(c) sodium
(d) gold
Answer:
(b) potassium

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Question 5.
Which one of the following is a more reactive element?
(a) Cu
(b) Li
(c) Zn
(d) Pb.
Answer:
(b) Li

Question 6.
Which one of the following is the least reactive element?
(a) Au
(b) Fe
(c) Ca
(d) Na.
Answer:
(a) Au

Question 7.
Double displacement reaction is also called as ______.
(a) Metastasis
(b) Metathesis
(c) Methanolysis
(d) Metalysis.
Answer:
(b) Metathesis

Question 8.
The acidic solution among the following is:
(a) seawater
(b) coffee
(c) lime water
(d) antacid
Answer:
(b) coffee

Question 9.
The reaction of sodium hydroxide with hydrochloric acid is an example for ______.
(a) Combination reaction
(b) Thermolysis reaction
(c) Neutralization reaction
(d) Precipitation reaction.
Answer:
(c) Neutralization reaction

Question 10.
In agriculture, the nature of the soil for rice is:
(a) alkaline
(b) neutral
(c) acidic
(d) none of the above
Answer:
(c) acidic

Question 11.
The role of manganese dioxide in the decomposition of potassium chlorate is ______.
(a) act as a reactant
(b) act as a catalyst
(c) act as a reagent
(d) act as a reaction medium.
Answer:
(b) act as a catalyst

Question 12.
The ionic product of water Kw is ______.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 11
Answer:
(c) \(\mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\)

Question 13.
Unit of ionic product of water is ______.
(a) mol-2 dm-6
(b) mol2 dm-6
(c) mol2 dm6
(d) mol dm-3.
Answer:
(b) mol2 dm-6

Question 14.
What is the value of ionic product of water?
(a) 1 × 10-4
(b) 1 × 10-1
(c) 1 × 1014
(d) 1 × 10-14.
Answer:
(d) 1 × 10-14.

Question 15.
pH of milk of magnesia is ______.
(a) 10
(b) 15
(c) 14
(d) 10.5.
Answer:
(a) 10

Question 16.
What is the relationship between pH and pOH?
(a) pH – pOH = 14
(b) pH + pOH = 14
(c) pH/pOH = 14
(d) pH + pOH = 1.4.
Answer:
(b) pH + pOH = 14

Question 17.
\(\mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2} \uparrow\)
The above reaction is an example of ______.
(а) Combination reaction
(b) Double displacement reaction
(c) Displacement reaction
(d) Decomposition reaction.
Answer:
(c) Displacement reaction

Question 18.
A student tests the pH of pure water using a pH paper. It shows a green colour. If a pH paper is after adding lemon juice to water, what colour will he observe?
(a) Green
(b) Red
(c) Yellow
(d) Pink.
Answer:
(b) Red

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Question 19.
When aqueous solution of silver nitrate and sodium chloride are mixed, ______ precipitate is immediately formed.
(a) white
(b) yellow
(c) red
(d) black.
Answer:
(a) white

Question 20.
pH = – log10 [H+]. The pH of a solution containing hydrogen ion concentration of 0.001 M solution is ______.
(a) 3
(b) 11
(c) 14
(d) 15.
Answer:
(a) 3

Question 21.
Silver anklet has got tarnished due to the formation of ______.
(a) Ag2O
(b) AgNO3
(c) Ag2S
(d) AgBr.
Answer:
(c) Ag2S

Question 22.
The colour of the precipitate formed by the reaction of lead nitrate with potassium iodide is ______.
(a) white
(b) yellow
(c) black
(d) red.
Answer:
(b) yellow

Question 23.
The reaction of calcium oxide with water is a ______ reaction.
(a) exothermic
(b) endothermic
(c) isothermic
(d) adiabatic.
Answer:
(a) exothermic

Question 24.
The gas released when calcium carbonate reacts with dilute hydrochloric acid is ______.
(a) NO2
(b) CO
(c) CO2
(d) H2.
Answer:
(c) CO2

Question 25.
The chemical used in whitewashing on the walls is ______.
(a) CaO
(b) CaCl2
(c) CaOCl2
(d) Ca(OH)2.
Answer:
(d) Ca(OH)2.

Question 26.
The chemical formula of marble is ______.
(a) CaCO3
(b) CaOCl2
(c) CaSO4. \(\frac { 1 }{ 2 }\) H2O
(d) Ca(OH)2.
Answer:
(a) CaCO3

Question 27.
Combustion of coal is an example of _____ reaction.
(a) displacement
(b) reduction
(c) decomposition
(d) combination.
Answer:
(d) combination.

Question 28.
The colour change takes place when copper carbonate is strongly heated is ______.
(a) green to black
(b) green to blue
(c) blue to green
(d) blue to black.
Answer:
(a) green to black

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Question 29.
The reaction of heat on copper carbonate into copper (II) oxide is _________ reaction.
(a) combination
(b) decomposition
(c) displacement
(d) redox.
Answer:
(b) decomposition

Question 30.
The dissolution of glucose in water is a _____ reaction.
(a) exothermic
(b) endothermic
(c) neutralisation
(d) combination.
Answer:
(b) endothermic

Question 31.
All combustion reactions are _______ reactions.
(a) combination
(b) exothermic
(c) endothermic
(d) neutralisation.
Answer:
(b) exothermic

Question 32.
The factor that affects the rate of the chemical reaction is ______.
(a) temperature
(b) concentration
(c) catalyst
(d) all the above.
Answer:
(d) all the above.

Question 33.
Magnesium ribbon reacts at a very faster rate with ______ acid.
(a) hydrochloric
(b) acetic
(c) formic
(d) oxalic.
Answer:
(a) hydrochloric

Question 34.
Our body metabolism is carried out by means of ______ secreted in our stomach.
(a) sulphuric acid
(b) hydrochloric acid
(c) nitric acid
(d) acetic acid.
Answer:
(b) hydrochloric acid

Question 35.
Which metals do not liberate gas on reaction with acids?
(a) Zn, Mg
(b) Ag, Cu
(c) Na, K
(d) Cr, Al.
Answer:
(b) Ag, Cu

Question 36.
When CO2 is passed through lime water, it turns ______.
(a) milky
(b) black
(c) red
(d) blue.
Answer:
(a) milky

Question 37.
The physical form of calcium carbonate is ______.
(a) limestone
(b) chalk
(c) marble
(d) all the above.
Answer:
(d) all the above.

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Question 38.
The colour change takes place when copper (II) oxide reacts with dilute hydrochloric acid is ______.
(a) blue to green
(b) black to green
(c) green to black
(d) green to blue.
Answer:
(b) black to green

Question 39.
Which of the following is a strong base?
(a) NH4OH
(b) Ca(OH)2
(c) Al(OH)3
(d) NaOH.
Answer:
(d) NaOH.

Question 40.
Zinc reacts with sodium hydroxide to form ______.
(a) Zinc hydroxide + H2O
(b) Sodium zincate + H2
(c) Zinc oxide + H2
(d) Zinc oxide + H2O.
Answer:
(b) Sodium zincate + H2

Question 41.
_____ reacts with sodium hydroxide.
(a) Cu
(b) Ag
(c) Cr
(d) Zn.
Answer:
(d) Zn.

Question 42.
________ is used as a medicine for stomach disorder.
(a) Sodium hydroxide
(b) Ammonium hydroxide
(c) Magnesium hydroxide
(d) Calcium hydroxide.
Answer:
(c) Magnesium hydroxide

Question 43.
The pH of stomach fluid is ______.
(a) 12
(b) 14
(c) 2
(d) 1.
Answer:
(c) 2

Question 42.
The hardest substance in the human body is ______.
(a) bone
(b) the enamel coating of teeth
(c) brain
(d) liver.
Answer:
(b) the enamel coating of teeth

Question 43.
Sugarcane requires _______ soil.
(a) acidic
(b) alkaline
(c) neutral
(d) amphoteric.
Answer:
(c) neutral

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Question 44.
Rice requires _____ soil.
(a) acidic
(b) basic
(c) alkaline
(d) neutral.
Answer:
(a) acidic

Question 45.
______ is a double salt.
(a) Sodium chloride
(b) Washing soda
(c) Potash alum
(d) Bleaching powder.
Answer:
(c) Potash alum

II. Fill in the blanks.

Question 1.
During chemical changes _____ are formed and these changes are more _____ than physical changes.
Answer:
New products, permanent.

Question 2.
Calcium oxide reacts with water to produce ______ and the reaction is ______.
Answer:
Slaked lime, exothermic.

Question 3.
During whitewashing, ______ reacts slowly with carbon dioxide in the air to form a thin layer of _______ on the walls.
Answer:
Calcium hydroxide, calcium carbonate.

Question 4.
When copper carbonate is heated, the products formed are _____ , ______ and change of colour from ______ to ____ is observed.
Answer:
CuO, CO2, green, black

Question 5.
When lead nitrate is heated, the gas liberated is ______ and its colour is ______.
Answer:
NO2, Reddish – brown

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Question 6.
Copper sulphate solution changes its blue colour into _____ colour when an iron nail is added to it and it acquires _____ colour.
Answer:
Green, brownish.

Question 7.
When Barium chloride reacts with sodium sulphate, the product formed is _____ and it is a ______ precipitate.
Answer:
Barium sulphate, white.

Question 8.
Powdered ____ reacts more quickly with hydrochloric acid than marble chips.
Answer:
CaCO3

Question 9.
Calcium carbonate present in the marble reacts with hydrochloric acid at a faster rate at _______ temperature.
Answer:
Higher.

Question 10.
______ is a substance which furnishes H+ ions when dissolved in water and a ______ is a substance which furnishes OH ions in water.
Answer:
Acid, Base.

Question 11.
Acids present in plants and animals are ______ and the acids in rocks and minerals are ______.
Answer:
Organic acid, inorganic acid.

Question 12.
Metal displaces _______ gas from dilute acid and the flame goes off with a ______ sound.
Answer:
Hydrogen, popping.

Question 13.
_____ is used in the manufacturing of soap and ______ is used in whitewashing buildings.
Answer:
Sodium hydroxide, calcium hydroxide.

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Question 14.
pH scale was introduced by ______ and the pH of the solution is 7, it is a ______ solution.
Answer:
S.P.L. Sorenson, neutral.

Question 15.
The pH of a normal healthy human skin is _____ and the pH of the stomach fluid is ______.
Answer:
7.4, 5.5.

Question 16.
The ideal pH for blood is _______ and pH of the mouth falls below ______.
Answer:
7.35 – 7.45, Sulphuric acid.

Question 17.
All photo decomposition reaction are _______ reactions.
Answer:
Endothermic.

Question 18.
Precipitation reactions give a _____ as the product.
Answer:
Insoluble salt.

Question 19.
Plants cannot grow in a ______ soil.
Answer:
Acidic.

Question 20.
Equilibrium is possible in a _____ system.
Answer:
Closed.

Question 21.
Pure water is a _______ electrolyte.
Answer:
Weak.

Question 22.
Most of the combination reactions are _______ in nature.
Answer:
Exothermic.

Question 23.
Silicon dioxide reacts with calcium oxide to form ______.
Answer:
Calcium silicate.

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Question 24.
Our mobile phones get energy from its _____ battery by chemical reaction.
Answer:
Lithium.

Question 25.
A ______ is a substance which increases the reaction rate.
Answer:
Catalyst.

Question 26.
pH range of human saliva is ______.
Answer:
6 – 8.

Question 27.
pH range of fresh milk is ______.
Answer:
5.

Question 28.
The pH of a solution can be determined by using a _______ indicator.
Answer:
Universal.

Question 29.
pH is ______.
Answer:
-log10 [H+].

Question 30.
The white enamel coating of our teeth is ______.
Answer:
Calcium phosphate.

III. Match the following.

Question 1.

(i) \(\mathrm{S}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{SO}_{2(\mathrm{g})}\) (a) Electrolytic decomposition reaction
(ii) \(2 \mathrm{NaCl}_{(\mathrm{aq})} \rightarrow 2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{g})}\) (b) Combination reaction
(iii) \(2 \mathrm{AgBr}_{(\mathrm{s})} \rightarrow 2 \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Br}_{2(\mathrm{g})}\) (c) Single displacement reaction
(iv) \(\mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{ZnCl}_{2(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}\) (d) Photo decomposition reaction

Answer:
i – b, ii – a, iii – d, iv – c.

Question 2.

Sample pH
(i) Oranges (a) 8
(ii) Grapes (b) 7
(iii) Pure water (c) 3.5
(iv) Egg white (d) 4

Answer:
i – c, ii – d, iii – b, iv – a.

Question 3.

Sample pH
(i) Sour milk (a) 4.2
(ii) Tomato juice (b) 4.5
(iii) Seawater (c) 12
(iv) Lime water (d) 8

Answer:
i – b, ii – a, iii – d, iv – c.

Question 4.

(i) Citrus fruits (a) Neutral soil
(ii) Rice (b) pH = 7
(iii) Sugarcane (c) alkaline soil
(iv) Rainwater (d) acidic soil

Answer:
i – c, ii – d, iii – a, iv – b.

Question 5.

(i) Sodium hydroxide (a) medicine
(ii) Calcium hydroxide (b) stain remover
(iii) Ammonium hydroxide (c) soap
(iv) Magnesium hydroxide (d) whitewashing

Answer:
i – c, ii – d, iii – b, iv – a.

Question 6.

(i) Gain of electrons (a) Oxidation reaction
(ii) Loss of electrons (b) Combination reaction
(iii) Combustion of coal (c) Redox reaction
(iv) Rusting of iron (d) Reduction reaction

Answer:
i – d, ii-a, iii – b, iv – e.

IV. State true or false. If false, give the correct statement.

Question 1.
Chemical changes are reversible changes.
Answer:
False.
Correct statement: Chemical changes are irreversible changes.

Question 2.
The silver anklet has got tarnished when exposed to air due to the formation of silver oxide. (Ag2O).
Answer:
False.
Correct statement: The silver anklet has got tarnished when exposed to air due to the formation of silver sulphide (Ag2S).

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Question 3.
Brisk effervescence takes place with the evolution of CO2 when calcium carbonate reacts with dilute hydrochloric acid.
Answer:
True.

Question 4.
The chemical formula for marble is CaCl2.
Answer:
False.
Correct statement: The chemical formula for marble is CaCO3

Question 5.
A reaction in which a single product is formed from two or more reactants is known as displacement reaction.
Answer:
False.
Correct statement: A reaction in which a single product is formed from two or more reactants is known as a combination reaction.

Question 6.
Lead Nitrate on heating decomposes to give lead oxide with the evolution of reddish-brown gas NO2 and O2 gas.
Answer:
True.

Question 7.
Any reaction that produces a precipitate is called a redox reaction.
Answer:
False.
Correct statement: Any reaction that produces a precipitate is called a precipitation reaction.

Question 8.
The reaction in which a more reactive element displaces a less reactive element from its compound is called displacement reaction.
Answer:
True.

Question 9.
The chemical reactions which take place with the evolution of heat energy are called endothermic reaction.
Answer:
False.
Correct statement: The chemical reactions which take place with the evolution of heat energy are called exothermic reaction.

Question 10.
All combustion reactions are endothermic reactions.
Answer:
False.
Correct statement: All combustion reactions are exothermic reactions.

Question 11.
Our body metabolism is carried out by means of sulphuric acid secreted in our stomach.
Answer:
False.
Correct statement: Our body metabolism is carried out by means of hydrochloric acid secreted in our stomach.

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Question 12.
Limestone, chalk and marble are different physical forms of calcium oxide.
Answer:
False.
Correct statement: Limestone, chalk and marble are different physical forms of calcium carbonate.

Question 13.
The atmosphere of earth is made up of thick white and yellowish clouds of sulphuric acid.
Answer:
False.
Correct statement: The atmosphere of Venus is made up of thick white and yellowish clouds of sulphuric acid.

Question 14.
Bond breaking releaser energy whereas bond formation absorbs energy.
Answer:
False.
Correct statement: Bond breaking absorbs energy whereas bond formation releases energy.

Question 15.
More active elements readily displace less active elements from their aqueous solution.
Answer:
True.

Question 16.
The irreversible reaction is relatively slow.
Answer:
False.
Correct statement: Irreversible reaction is fast.

V. Assertion and Reason

Question 1.
Assertion (A): The lustrous white colour of the silver anklet slowly changes into slightly back colour.
Reason (R): silver anklet reacts with H2S in the air to form Ag2S (Silver Sulphide) which is black in colour.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) Both (A) and (R) are wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

SamacheerKalvi.Guru

Question 2.
Assertion (A): The reaction of calcium oxide with water is an exothermic reaction.
Reason (R): The reaction is accompanied by a hissing sound and formation of bubbles leading to the absorption of a considerable amount of heat.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 3.
Assertion (A): During the reaction of calcium carbonate with dilute hydrochloric acid, brisk effervescence takes place.
Reason (R): Brisk effervescence is due to the evolution of carbon dioxide gas.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 4.
Assertion: When copper carbonate is heated strongly, the green colour is changed into a black colour.
Reason (R): The colour change is due to the decomposition of copper carbonate into copper oxide.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 5.
Assertion (A): when lead nitrate is heated, the gas released has red – orange colour and it is lead oxide.
Reason (R): Lead nitrate on heating undergoes combination reaction.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are wrong

Question 6.
Assertion (A): iron js more reactive than copper.
Reason (R): iron js displaced from iron sulphate by copper.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(b) (A) is correct but (R) is wrong

Question 7.
Assertion (A): Copper displaces zinc (or) lead from the salt solution.
Reason (R): Copper is more reactive than zinc and lead.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(d) Both (A) and (R) are wrong.

Question 8.
Assertion (A): All combustion reactions are exothermic reactions.
Reason (R): During combustion reaction, heat energy is liberated.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(a) Both (A) and (R) are correct

SamacheerKalvi.Guru

Question 9.
Assertion (A): when glucose is kept on our tongue, a cooling effect is felt.
Reason (R): It is an endothermic reaction in which heat is absorbed.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): Powdered calcium carbonate reacts more quickly with hydrochloric acid than marble chips.
Reason (R): Powdered calcium carbonate offers a large surface area than marble chips. Because the greater the surface area, the greater is the rate of the reaction.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 11.
Assertion (A): Combination reaction is exothermic in nature.
Reason (R): During the formation of new bonds, releases a huge amount of energy in the form of heat.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) does not explain (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) (A) is correct and (R) explains (A)

Question 12.
Assertion (A): When glucose is kept on our tongue, a cooling effect is felt.
Reason (R): It is an endothermic reaction in which heat is absorbed.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) does not explain (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) (A) is correct and (R) explains (A)

VI. Short Answer Questions.

Question 1.
What type of chemical reaction takes place when
(i) limestone is heated
(ii) a magnesium ribbon is burnt in air?
Answer:
(i) When limestone is heated, the reaction takes place is a decomposition reaction.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 12

(ii) When a magnesium ribbon is burnt in air, it is an oxidation reaction.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 13

Question 2.
What is a combustion reaction?
Answer:
A combustion reaction is one in which the reactant rapidly combines with oxygen to form one or more oxides and energy (heat). So in combustion reactions, one of the reactants must be oxygen.

Question 3.
Which of the following is a combustion?

  1. Digestion of food
  2. Rusting of iron.

Answer:

  1. Digestion of food : Not a combustion reaction, because it is a endothermic process and where energy is utilized.
  2. Rusting of iron : Combustion reaction.

Question 4.
When the lead powder is added to copper chloride solution, a displacement reaction occurs and solid copper is formed.
(i) Write the equation for the reaction.
(ii) Why does the displacement reaction occur?
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 14

(ii) Copper is less reactive than lead. So lead has displaced copper from copper chloride. It is a displacement reaction.

Question 5.
What happens when lead nitrate reacts with potassium iodide solution?
Answer:
Lead nitrate solution when reacts with a potassium iodide solution, a deep yellow precipitate of Pbl2 is formed.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 15

Question 6.
Differentiate Exothermic reaction and Endothermic reaction.
Answer:

Exothermic reaction Endothermic reaction
1. The chemical reactions which take place with the evolution of heat energy are called exothermic reactions. 1. The chemical reactions which take place with the absorption of heat energy a called endothermic reactions.
2. e.g. All combustion reactions are exothermic. 2. e.g. Dissolution of glucose is the endothermic reaction.
3. N2 + 3H2 → 2NH3 + Heat 3. 2NH3 + Heat → N2 + 3H2

Question 7.
Why the study of reaction rate is important?
Answer:
Faster the reaction, more will be the amount of the product in a specified time. So, the rate of a reaction is important for a chemist for designing a process to get a good yield of a product. Rate of reaction is also important for a food processor who hopes to slow down the reactions that cause food to spoil.

Question 8.
What is meant by decomposition reaction? Give an example.
Answer:
A single compound breaks down to produce two or more substances. Such type of reaction is called decomposition reaction.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 16

Question 9.
Give reason.
(a) Granulated Zinc reacts faster with 2M HCl than 1M HCl.
Answer:
Reason: As the concentration of the reactant increases the rate of the reaction increases.

(b) Food kept at room temperature spoils faster than that kept in the refrigerator.
Answer:
Reason: In the refrigerator the temperature is lower than room temperature, so the reaction rate is less.

Question 10.
What is meant by double displacement reaction? Give an example.
Answer:
A double decomposition reaction is a reaction in which the exchange of ions between two reactants occur leading to the formation of two different products.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 17

Question 11.
Why the toothpastes are generally basic in nature?
Answer:
The pH of the saliva is usually between 6.5 to 7.5, when the pH of the mouth saliva falls below 5.5 the enamel coating of our teeth calcium phosphate gets weathered. That is why the tooth pastes are generally basic in nature.

Question 12.
What happens to food containing fat and oil kept open for a long time?
Answer:
When food containing fat and oil is left as such for a long time, it becomes stale. The stale food develops bad taste and foul smell. This is very common in curd and cheese, particularly in summer. Oils and fats are slowly oxidised to certain foul – smelling compounds.

Question 13.
Define the rate of a chemical reaction.
Answer:
The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or product per unit time.
\(\text { Rate }=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\).

Question 14.
Define catalyst.
Answer:
A substance which alters the rate of a reaction without undergoing any change in mass and composition is known as a catalyst.

SamacheerKalvi.Guru

Question 15.
What happens during a chemical reaction?
Answer:

  • In a chemical reaction, the atoms of the reacting molecules or elements are rearranged to form new molecules.
  • Old chemical bonds between atoms are broken and new chemical bonds are formed.
  • Bond breaking absorbs energy whereas bond formation releases energy.

Question 16.
What; is a balanced chemical equation?
Answer:
A balanced chemical equation is the simplified representation of a chemical reaction which describes the chemical composition, the physical state of the reactants and the products, and the reaction conditions.

Question 17.
What are the main classes of decomposition reactions?
Answer:
There are three main classes of decomposition reactions. They are,

  • Thermal decomposition reactions
  • Electrolytic decomposition reactions
  • Photo decomposition reactions

Question 18.
What are thermal decomposition reactions?
Answer:
In a thermal decomposition reaction, the reactant is decomposed by applying heat.
For example, on heating mercury (II) oxide is decomposed into mercury metal and oxygen gas.
\(2 \mathrm{HgO}_{(\mathrm{s})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Hg}_{(\mathrm{l})}+\mathrm{O}_{2(\mathrm{g})}\).

Question 19.
What are Electrolytic decomposition reactions?
Answer:
In this type of reaction, the reactant is decomposed by applying electricity.
For example, decomposition of sodium chloride occurs on passing electric current through its aqueous solution.
\(2 \mathrm{NaCl}_{(\mathrm{aq})} \stackrel{\text { Electricity }}{\longrightarrow} 2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{g})}\).

Question 20.
What is the photodecomposition reaction?
Answer:
In this type of reaction, the reactant is decomposed by applying light.
For example, when silver bromide is exposed to light, it breaks down into silver metal and bromine gas.
\(2 \mathrm{AgBr}_{(\mathrm{s})} \stackrel{\mathrm{Light}}{\longrightarrow} 2 \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Br}_{2(\mathrm{g})}\).

Question 21.
What is Metathesis reaction?
Answer:
The ion of one compound is replaced by the ion of another compound. Ions of identical charges are only interchanged, i.e. a cation can be replaced by another cation. This reaction is called metathesis reaction.

Question 22.
What is a precipitation reaction?
Answer:
When aqueous solution of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.
e.g„ \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{aq})}+2 \mathrm{KI}_{(\mathrm{aq})} \longrightarrow \mathrm{PbI}_{2(\mathrm{s})} \downarrow+2 \mathrm{KNO}_{3(\mathrm{aq})}\).

Question 23.
How will you distinguish between combination and decomposition reactions?
Answer:

Combination Reactions Decomposition Reactions
One or more reactants combine to form a single product A single reactant is decomposed to form one or more products
Energy is released Energy is absorbed
Elements or compounds may be the reactants The single compound is the reactant

Question 24.
What is a neutralization reaction?
Answer:
It is another type of displacement reaction in which the acid reacts with the base to form a salt and water. It is called neutralization reaction as both acid and base neutralizes each other.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 18

Question 25.
What is Combustion reaction?
Answer:
A combustion reaction is one in which the reactant rapidly combines with oxygen to form one or more oxides and energy (heat).
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 19

Question 26.
What are reversible and irreversible reactions?
Answer:
Reversible reaction : A reversible reaction is a reaction that can be reversed, i.e., the products can be converted back to the reactants.
e.g., \(\mathrm{PCl}_{5(\mathrm{g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})}\)
Irreversible reaction: The reaction that cannot be reversed is called irreversible reaction. The irreversible reactions are unidirectional.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 20

Question 27.
Why is the reaction rate important?
Answer:
Faster the reaction, more will be the amount of the product in a specified time. So, the rate of a reaction is important for a chemist for designing a process to get a good yield of a product. Rate of reaction is also important for a food processor who hopes to slow down the reactions that cause food to spoil.

VII. Long Answer Questions.

Question 1.
Suggest a reason for each observation given below.
Answer:
(i) In fireworks, powdered Mg is used rather than Mg ribbon.
Powdered Mg will have larger surface area than Mg ribbon and the rate of the reaction increases.

(ii) Zn and dil. H2SO4 react much more quickly when a few drops of CuSO4 solution are added.
When few drops of CuSO4 are added to the solution containing Zn and dil. H2SO4, the rate of the reaction increases, because CuSO4 acts as catalyst.

(iii) The reaction between MgCO3 and dil. HCl speeds up when some con. HCl is added. As the concentration of the reactant increases, the rate of the reaction increases.

Question 2.
Observe the given chemical change and answer the following:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 21

  1. Identify ‘A’ and ‘B’.
  2. Write the commercial name of calcium hydroxide.
  3. Identify products ‘C’ and ‘D’ when HCl is allowed to oxide react with calcium oxide.
  4. Say whether calcium oxide is acidic or basic.

Answer:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 22

  1. A is calcium carbonate. B is carbon-di-oxide.
  2. Slaked lime is the commercial name of calcium hydroxide.
  3. The products C and D are calcium chloride (CaCl2) and water (H2O).
  4. Calcium oxide is basic in nature.

Question 3.
Take copper nitrate in a test tube and heat it over the flame.

  1. What is the colour of cupric nitrate?
  2. What do you observe?
  3. Name the type of reaction that takes place.
  4. Write the balanced equation.

Answer:

  1. The colour of cupric nitrate is Blue.
  2. When cupric nitrate is heated in a test tube, we can observe the evolution of reddish-brown gas (NO2) Nitrogen dioxide.
  3. The reaction takes place is a decomposition reaction.
  4. Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 23

Question 4.
Redox reactions are reactions during which electron transfer takes place. Here magnesium atom transfers two electrons one each to the two chlorine atoms.
(i) What are the products of this reaction?
(ii) Write the balanced equation for the complete reaction.
(iii) Which element is being oxidized?
(iv) Which element is being reduced?
(v) Write the reduction part of the reaction.
Answer:
(i) Magnesium atom is converted to Mg2+ ion. Two chlorine atoms are converted to 2Cl ions. So the products are Mg2+ ion and 2Cl ions.
(ii)
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 24
(iii) Mg atom is being oxidised by donating 2 electrons.
(iv) Cl2 molecule is being reduced by accepting 2 electrons.
(v) Cl2 + 2e → 2Cl. It is a reduction reaction in which gain of e take place.

Question 5.
Take Cu(NO3)2 in a test tube and heat it over the flame.
(i) What is the colour of Cu(NO3)2?
Answer:
Blue

(ii) What do you observe?
Answer:
Evolution of reddish-brown gas.

(iii) Name the type of reaction that takes place.
Answer:
Decomposition.

(iv) Write the balanced equation.
Answer:
Cu(NO3)2 → 2CuO + 4NO2 + O2

Question 6.
Sodium hydroxide and hydrochloric acid react as shown in this equation.
\(\mathbf{N a O H}_{(\mathrm{aq})}+\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathbf{H}_{2} \mathbf{O}_{(\mathrm{l})}\)
(i) Which type of chemical reaction is this?
(ii) The reaction is exothermic. Explain what that means.
(iii) Differentiate exothermic reaction and endothermic reaction.
(iv) What happens to the temperature of the solution as the chemicals react?
Answer:
(i) It is a neutralisation reaction.
(ii) An exothermic reaction is a chemical reaction in which the evolution of heat energy takes place.
(iii)

Exothermic reaction Endothermic reaction
1. The reaction In which heat energy is evolved is known as an exothermic process. 1. The reaction in which heat energy is absorbed is known as an endothermic process.
2. N2 + 3H2 → 2NH3 + Heat 2. 2NH3 + Heat → N2 + 3H2

(iv) When NaOH reacts with HCl to give NaCl and water, heat is evolved. So the solution’s temperature increases.

VIII. HOT Questions.

Question 1.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid is added to test tube A, while acetic acid is added to test tube B. The amount and concentration taken for both the acids are same. In which test tube does the reaction occur more vigorously and why?
Answer:
In the test tube, the reaction occurs more vigorously.
Comparing hydrochloric acid and acetic acid, HCl (hydrochloric acid) is a strong acid and more reactive whereas acetic acid is a weak organic acid and less reactive.
\(\mathrm{Mg}+2 \mathrm{HCl} \stackrel{\text { Fast reaction }}{\longrightarrow} \mathrm{MgCl}_{2}+\mathrm{H}_{2} \uparrow\).

Question 2.
Classify the following reactions based on the rate of the reactions as very fast or instantaneous slow and moderate reactions.
(a) AgNO3(aq) + NaCl(aq) → AgCl ↓ + NaNO3(aq)
Answer:
Very fast reaction (or) Instantaneous reaction.

(b) 2 Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Answer:
Very fast reaction.

(c) Rusting of iron.
Answer:
Very slow reaction.

(d) Inversion of cane sugar into Glucose and fructose.
Answer:
Moderately slow reaction.

(e) Fermentation of sugar into alcohol.
Answer:
Very slow reaction.

Question 3.
What is the chemical reaction taken place in the tarnishing of silver anklet?
Answer:
The lustrous white colour of the silver anklet slowly changes into a slightly black colour. It is due to the formation of silver sulphide (Ag2S) as a result of the reaction between silver and hydrogen sulphide in the air.

Question 4.
Why toothpaste are basic in nature?
Answer:
The white enamel coating of our teeth is calcium phosphate, the hardest substance in our body. It does not dissolve in water. If the pH of mouth falls below 5.5, the enamel gets corroded. Toothpaste is generally basic and used for cleaning the teeth can neutralize the excess acid and prevent tooth decay.

Question 5.
Why the solution of slaked lime is used for whitewashing?
Answer:
A solution of slaked lime is used for whitewashing walls. Calcium hydroxide reacts slowly with the carbon dioxide in the air to form a thin layer of calcium carbonate on the walls. Calcium carbonate is formed after two to three days of whitewashing and gives a shiny finish to the walls. It is interesting to note that the chemical formula for marble is also CaCO3.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 25

Question 6.
Complete the following reactions.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 26
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 27

Question 7.
Let us consider the following two reactions.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 28
Which reaction will not occur. Why?
Answer:
The first reaction involves the displacement of chlorine from NaCl, by fluorine. In the second reaction, chlorine displaces fluorine from NaF. Out of these two, the second reaction will not occur. Because fluorine is more active than chlorine and occupies the upper position in the periodic table. So, in displacement reactions, the activity of the elements and their relative position in the periodic table are the key factors to determine the feasibility of the reactions. More active elements readily displace less active elements from their aqueous solutions.

Question 8.
Which of the metals displaces hydrogen gas from hydrochloric acid? Silver or zinc. Give the chemical equation of the reaction and justify your answer.
Answer:
Zinc displaces hydrogen gas from hydrochloric, acid.
Zinc is more reactive than silver.
Zn + 2HCl → ZnCl2 + H2 ↑.

Question 9.
Foods kept at room temperature spoils faster than that kept in the refrigerator. Why?
Answer:
Food kept at room temperature spoils faster than that kept in the refrigerator. In the refrigerator, the temperature is lower than the room temperature and hence the reaction rate is less.

Question 10.
How will you enhance the rate of decomposition of potassium chlorate?
Answer:
On heating potassium chlorate, it decomposes into potassium chloride and oxygen gas, but at a slower rate. If manganese dioxide is added, it increases the reaction rate.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 29
Here MnO2 is the catalyst. Therefore the addition of MnO2 enhances the rate of decomposition of potassium chlorate.

Question 11.
Powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips. Why?
Answer:
Powdering of the reactants increases the surface area and more energy is available on the collision of the reactant particles. Thus, the reaction rate is increased. Hence powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips.

Samacheer Kalvi 10th Science Types of Chemical Reactions Additional Problems Solved

Question 1.
The [OH] ion concentration of a solution is 1.0 × 10-8 M. What is the pH of the solution?
Solution:
The concentration of hydroxide ion = [OH] = 1.0 × 10-8 M.
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 30

Question 2.
The hydrogen ion concentration of a solution is 1.0 × 10-9 m. What is the pH of the solution? Find out whether the given solution is acidic, basic or neutral.
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 31

Question 3.
The hydroxide ion concentration of a solution is 0.001 m. What is the pH of the solution?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 32
pH = 14 – pOH
pH = 14 – 3 = 11
pH = 11.

Question 4.
The hydroxide ion concentration of a solution is 1.0 × 10-9 m. What is the pH of the solution?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 33

Question 5.
The hydrogen ion concentration of a solution is 1 × 10-4 m. Calculate the pH and pOH of that solution.
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 34

Question 6.
Calculate the pH of sodium hydroxide solution having the concentration of OH 0.01 mL-1.
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions 35

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Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Mapping Skills Textual Exercise

I. Choose the correct answer.

Question 1.
The new phase in topographical surveying in the 20th century is ……
(a) toposheets
(b) aerial photography
(c) maps
(d) satellite imagery
Answer:
(d) satellite imagery

Question 2.
…… indicates the purpose or theme of the map.
(a) Title
(b) Scale
(c) Direction
(d) Legend
Answer:
(a) Title
Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 3.
Standard symbols that are used in maps to convey a definite meaning are called ………
(a) conventional signs and symbols
(b) coordinates
(c) grid references
(d) directions
Answer:
(a) conventional signs and symbols

Question 4.
Which one of the following maps show us a very large area with less details?
(a) Large scale
(b) Thematic
(c) Physical
(d) Small scale
Answer:
(d) Small scale

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 5.
GPS consists of a constellation of ……… satellites.
(a) 7
(b) 24
(c) 3.2
(d) 64
Answer:
(b) 24

II. Consider the given statements and choose the right option given below.

Question 1.
Assertion(A): The points at which the vertical and horizontal lines of the grid intersect are called coordinates.
Reason(R): The lines that run horizontally and vertically are called Northings and Eastings respectively.
(a) Both (A) and (R) are true ; (R) explains (A)
(b) Both (A) and (R) are true ; (R) does not explain (A)
(c) (A) is correct; (R) is false
(d) (A) is false ; (R) is true
Answer:
(a) Both (A) and (R) are true ; (R) explains (A)

Question 2.
Assertion(A): The legend of a map does not help us to understand the information in a map. Reason(R): It is usually placed at the left or right comer at the bottom of the map.
(a) (A) is false ; (R) is true
(b) Both (A) and (R) are true ; (R) does not explain (A)
(c) (A) is correct; (R) is false
(d) Both (A) and (R) are true ; (R) explains (A)
Answer:
(a) (A) is false ; (R) is true

III. Match the following:

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills 1
Answers:
1. (e)
2. (d)
3. (b)
4. (c)
5. (a)

IV. Answer in brief

Question 1.
Name the different methods to represent the Earth.
Answer:

  1. A map is the basic tool of a geographer. It illustrates the earth’s surface clearly and effectively through a combination of drawings, words, and symbols.
  2. A map projection is a way of showing the spherical-shaped earth on a flat piece of paper.
  3. A map projection is a systematic transformation of the latitudes and longitudes of location from the surface of a sphere or an ellipsoid into locations on a plane.
  4. A Globe is a spherical model of earth. Globe Serve similar purposes to maps but unlike maps do not disturb the surface that they portray except to scale it down. A globe of the earth is called a terrestrial globe.

Question 2.
What is a map?
Answer:
A map is the basic tool of a geographer. It illustrates the earth’s surface Clearly and effectively through a combination of drawings, words, and symbols. A map is a location guide.

Question 3.
What are the components of a map?
Answer:
A map should include the following components namely, the title, scale, direction, grid reference, projection, legend, conventional signs, and symbols.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 4.
The distance between the two cities A and B is 5 km. It is represented by a line of 5 cm on the map. Calculate the distance and give the answer in RF.
Answer:
Representative Fraction (R.F.) = Distance on the map / Distance on the ground Given, Distance on the map = 5 cm
The distance on the ground = 5 km
∴ R.F. = 5 cm / 5 km
Converting km to cm; 5 km = 500000 cm
So R.F. is 5 : 500000
i.e. R.F. is 1 : 100000

Question 5.
Mention a few surveying instruments.
Answer:
Geographers mainly use Chain, Prismatic compass, Plane table, Dumpy level, Abney level, Clinometer, Theodolite, Total Station, and GNSS to measure the distance, angle, altitude, and position of the area of survey.

Question 6.
Define remote sensing.
Answer:
Remote Sensing refers to the observation and measurement of earthly objects without touching them. ‘Remote’ means far away and ‘Sensing’ means observing or collecting information. Remote sensing means acquiring information of things/places from a distance, using a variety of tools and methods.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 7.
What are the components of remote sensing?
Answer:
Components of remote sensing are

  1. Energy source
  2. Transmission path
  3. Target
  4. Sensor

V. Give Reasons

Question 1.
Satellite imageries stimulate map making.
Answer:

  1. Satellite imagery refers to digitally transmitted images of the satellites. Therefore it can be easily integrated with software for the improvement of images.
  2. Satellites circle the Earth or remain geostationary and therefore, changes in weather or any other natural or man-made modifications do not affect the functioning of Satellites.

Question 2.
A map is the basic tool of a geographer.
Answer:
With maps on hand, one can see the world in one sweep. A map is worth a thousand words. Maps are introduced with its components such as scale, signs and symbols. A map is the basic tool of a geographer. It illustrates the earth’s surface clearly and effectively through a combination of drawings, words and symbols. A map is a location guide.

Question 3.
Grid references are essential to finding the exact location of places on a map.
Answer:

  1. The location of a place can be simply defined by its latitude and longitude.
  2. The points at which the vertical and horizontal lines of the grid intersect are called coordinates.
  3. Therefore grid references are essential to finding the exact location of a place.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 4.
Web cartography is one of the modern mapping techniques.
Answer:
The introduction of web mapping is a major new trend in cartography. The term Web cartography is connected with the design, production, display, and use of maps over the Web. Various types of maps are present on the Web. Until recently cartography was restricted as it required relatively expensive and complex hardware and software as well as skilled cartographers and geomatics engineers.

Web-based GIS has brought many geographical datasets, including free ones generated by OpenStreetMap and proprietary datasets owned by Navteq, Google, Waze, and others. A range of free software to generate maps has also been conceived and implemented alongside proprietary tools like ArcGIS. As a result, the barrier to entry for serving maps on the web has been lowered.

VI. Distinguish Between The Following:

Question 1.
Globe and Map
Answer:

Globe Map
It is a representation of the whole earth with the clear marking of longitude and latitude of all the places on earth. A map illustrates the earth’s surface more clearly and effectively through a combination of drawings, words and symbols.

Question 2.
Large scale map and small scale map
Answer:

Large scale map Small scale map
Large amount of detail is shown of a small area can be seen in a large-scale map. Small amount of detail of a larger area can be seen in a small scale map.

Question 3.
Aerial photographs and satellite imageries
Answer:

Aerial photographs Satellite imageries
It covers a small area and needs permission from the authorities. It allows global coverage and does not require permission.
Revisits or repeatability involves extra cost. Satellites circle the Earth; they can repeat and revisit easily.

Question 4.
GIS and GPS
Answer:

GIS GPS
Geographic Information System is a computer-based tool for managing a large amount of data collected for a given geographic region through remote sensing, GPS and other sources. GPS is the U.S. implementation of the world’s first and currently the most used Global Navigation Satellite System (GNSS) created by the U. S. Department Of Defense.
GIS finds its strongest use in resources management, telecommunications and urban and regional planning. Weather forecasting, earthquake monitoring and environmental protection can be done effectively by using GPS.

VII Answer in a paragraph:

Question 1.
What do you mean by the term ‘scale of the map’? Explain its classification.
Answer:

  1. Scale is one of the components of a map.
  2. Scale makes it possible to reduce the size of the whole earth to show it on a piece of paper.
  3. A scale is a ratio between the actual distance on the map to the actual distance on the ground.
  4. Scales can be represented in three methods. They are
    • Statement
    • Representative Fraction (R.F)
    • Linear or Graphical scale methods.

(a) Statement Scale:
The statement scale describes the relationship of map distance to the ground distance in words, such as one centimetre to ten kilometres. It is expressed as 1cm = 10 km.

(b) The Representative Fraction (R.F):

  1. It describes the proportion or ratio of the map distance to ground distance. It is usually abbreviated as R.F. It is stated as 1/100000 or 1:100000
  2. This means that one unit on the map represents 100,000 of the same unit on the ground.
  3. This unit may be an inch or a centimetre or any other linear measurement unit.
    Representative Fraction (R.F.) = \(\frac { Distance on the map }{ Distance on the Ground }\)

(c) Linear (or) Graphical scale: In geography, a linear scale is represented by a straight line divided into equal parts (Primary and secondary) to show what these markings represent on the actual ground. This scale helps in the direct measurement of distance on the map.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 2.
Write a note on directions with a relevant diagram.
Answer:
Direction
Maps are drawn normally with north orientation. The North direction in a map is always towards the North Pole of the earth. If you position yourself looking at the North Pole, on your right will be the east; your left will be the west; at your back will be south. These four main directions are called the cardinal directions. The direction is usually indicated on a map by a North-South line, with the North direction represented by an arrow head.
Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills 80

Question 3.
What are the three major functional segments of GPS? Explain about anyone.
Answer:
GPS has made a considerable impact on almost all positioning, navigation, timing and monitoring applications. It provides particularly coded satellite signals that can be processed in a GPS receiver, allowing the receiver to estimate position, velocity and time.

  1. The (GPS) Global Positioning System is a U.S. owned utility that provides users with positioning, control segment and the user segment.
  2. The GPS ground segment (also referred to as control segment or operational control system) is responsible for the proper operation of the GPS system.
  3. The GPS control segment is composed of network of monitor stations (MS), a Master Control Systems (MCS) a backup of the MCS and Ground Antenna (GA).
  4. The GPS space segment consists of a constellation of transmitting radio signal to users. The linked states is committed to maintaining the availability of atleast 24 operational GPS satellites, 95% of the times.
  5. The user segment is the practice of dividing all customers into segments based on characteristics they share. For example sorting users by region, language, or behaviour.

Question 4.
Bhuvan has tremendous uses for scientists, policymakers, or the general public. Justify.
Answer:
Bhuvan (Sanskrit for Earth) is a free internet-based computer application launched by the Indian Space Research Organization (ISRO) on August 12th, 2009. It enables visualization of Indian Remote Sensing (IRS) images taken over a year ago, by ISRO’s seven satellites, including CartoSat-1 and CartoSat-2. Using Bhuvan connected to the Internet, one can explore places of interest, scenes of events in the news or parts around the world they may never visit in person, by either entering the names of places or coordinates (latitudes and longitudes). Bhuvan has tremendous uses for scientists, academicians, policymakers, and the general public.

Advantages

  • Bhuvan, due to 3D rendering, gives the impression of moving through real space through the entire globe
  • Students can use Bhuvan to understand subjects ranging from Sciences to History of places.
  • It provides information on natural resources and timely information on disasters.
  • Administrators use it for monitoring various developmental schemes.

VIII. HOTS:

Question 1.
Can you imagine a world without satellites?
Answer:

  1. Today a growing number of satellites orbit around the Earth, making various earth observation, communication, navigation, and science application possible.
  2. While we may not always realize or acknowledge their existence, the important role these systems play in our daily lives cannot be underestimated.
  3. As technology advances, the potential of satellites will undoubtedly continue to grow. New markets will merge along with new opportunities to push the boundaries of what space technology currently offers.
  4. Likely satellites reach just about everywhere today. In certain isolated parts of our planet. Satellites provide inhabitants with access to telephone service, 4G / 5G, broadband, and work.
  5. GPS helps us to travel from A to B without getting lost along the way.

Question 2.
Imagine you are a cartographer. Plan and bring out a few ideas to improve your area.
Answer:
Cartographers are trained in aerial photography and in collecting survey data for preparing maps, charts and sketches.

They work to create detailed information maps based on survey data.
Ideas to improve Chennai

  1. Maintaining the ecological process in Chennai.
  2. Balancing the essential infrastructures of the urban areas.
  3. Improving the connectivity in and around South Chennai from the main city.
  4. Improving Civic Services

Chennai Corporation is set to upgrade its Geographical Information System (GIS) services. This was used for the creation of a digital base and Utility mapping using GIS technology in 44,000 streets in Chennai.
(Students can add their own creative ideas)

IX Map skill:

A. With the help of an atlas, mark the following on the outline map of Tamil Nadu.
a. The latitude and longitude of Chennai. .
b. Mark the city located at 10° N, 78° E.
c. Locate the city approximately on 11°N and 76°E.
d. Find the latitude and longitude of Kanyakumari and mark it.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills 85

Mapping Skills Additional Questions

I. Choose the correct answer.

Question 1.
The first ancient Greek to draw a map of the Known World is ………
(a) Anaximander
(b) Gerardus Mercator
(c) Felix Nadar
(d) None of these
Answer:
(a) Anaximander

Question 2.
The foundation for map-making in India was laid during the …… period.
(a) Pre-historic
(b) Medieval
(c) Vedic
(d) Modern
Answer:
(c) Vedic

Question 3.
……. is a way of showing the spherical shaped earth on a flat piece of paper.
(a) Direction
(b) Grid Reference
(c) Projection
(d) Legend
Answer:
(c) Projection

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 4.
……… symbols are used to indicate railways, roads, power lines, telephone lines in
mapping.
(a) Line
(b) Point
(c) Area
(d) None of these
Answer:
(a) Line

Question 5.
Maps produced by analysis can be used to pinpoint problem areas.
(a) GPS
(b) GIS
(c) GNSS
(d) DoD
Answer:
(b) GIS

II. Find out the correct statement.

Question 1.
Assertion (A): A linear scale is represented by a straight line divided into equal parts to show what these markings represent on the actual ground.
Reason (R): This scale helps in the direct measurement of distance on the map.
(a) Both A and R are true; R explains A
(b) Both A and R are true; R does not explain A
(c) A is correct and R is wrong
(d) A is wrong and R is correct
Answer:
(a) Both A and R are true; R explains A

Question 2.
Assertion (A): Satellites do not collect large amounts of data of the entire area in a short span. Reason (R): Sensors in the satellites record the reflected and emitted radiation.
(a) A is wrong and R is correct
(b) Both A and R are hue; R does not explain A
(c) A is correct and R is wrong
(d) Both A and R are true; R explains A
Answer:
(a) A is wrong and R is correct

III. Match the following: 

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills 25
Answers:
1. (e)
2. (a)
3. (b)
4. (c)
5. (d)

IV. Answer in brief.

Question 1.
Mention the first mapmaker.
Answer:
Anaximander was the first ancient Greek to draw a map of the known world. It is for this reason that he is considered by many to be the first mapmaker (the first cartographer).

Question 2.
Who is a Cartographer? What is Cartography?
Answer:
A cartographer is one who measures, analyzes, and interprets geographical information to create maps and charts for political, cultural, and educational purposes. The art and science of map-making are called Cartography.

Question 3.
What do you mean by ‘Direction’?
Answer:
Maps are drawn normally with north orientation. The North direction in a map is always towards ‘ the North Pole of the earth. If you position yourself looking at the North Pole, on your right will be the east; your left will be the west; at your back will be south. These four main directions are called the cardinal directions. The direction is usually indicated on a map by a North-South line, with the North direction represented by an arrowhead.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills

Question 4.
What is Projection? Mention the three methods in widest use.
Answer:
A map projection is a way of showing the spherical shaped earth on a flat piece of paper.
The three methods in widest use are as follows:

  • Projection on the surface of a cylinder
  • Projection on to the surface of a cone
  • Projection directly onto a flat plane called planar or zenithal or azimuthal projection

Question 5.
Remote sensing can be either passive or active – Explain.
Answer:

  1. Remote sensing can be either passive or active.
  2. Active systems emit their own source of light energy such as RADAR.
  3. Passive systems depend on sunlight as an energy source.

Question 6.
What are the types of Maps?
Answer:
(a) On the basis of scale, maps can be classified into:

  • Large scale maps: A large amount of detail; can only show a small area.
  • Small scale map: Small amount of detail; can show a large area.

(b) On the basis of utility and purpose, they are classified as:

  • General maps / Topographic Maps (physical and political maps)
  • Thematic map (spatial variations of single phenomena)
  • Special purpose maps (Braille maps for blind people, maps for neo literates, military maps, navigational charts, etc).

Question 7.
What are the geo objects?
Answer:

  1. Geographical objects in the real world are matched to program objects known as geo objects.
  2. Geo objects include placemarks, circles, polylines, rectangles, polygons, and their collections.
  3. Place markers indicate a place on a map.

V. Give reasons

Question 1.
The foundation for map-making in India was laid during the Vedic period.
Answer:
The foundation for map-making in India was laid during the Vedic period. Mahabharata conceived a round world surrounded by water. Surveying and map-making were an integral part of the revenue collection procedure in the medieval period. E.g.: Sher Shah Suri’s revenue maps and Rajendra Chola’s land survey techniques.

Question 2.
Conventional signs and symbols are standard symbols used on a map.
Answer:

  1. A map is a global language and it needs to be drawn according to the international standards
  2. Conventional signs and symbols are standard symbols used on a map and explained in the legend to convey a definite meaning.

Question 3.
“GPS helps in providing accurate transport data”.
Answer:

  1. GPS technology has tremendous applications in everything.
  2. It helps in military searches and rescue in wars.
  3. It can work as a reliable tourist guide (distance, route and direction)

VI. Distinguish between the following.

Question 1.
Maps and Cartography
Answer:

Maps Cartography
Maps are drawings of an area as seen from above. Maps can show a whole or part of the world. The art and science of map-making are called cartography.

VII. Answer in a paragraph.

Question 1.
What is Satellite Remote Sensing? Explain the
(a) components of Remote Sensing
(b) process of Remote Sensing.
Answer:
Satellite remote sensing is the science of collecting data about an object or area from artificial satellites orbiting the Earth. The term ‘satellite imagery’ refers to digitally transmitted images of the satellites.
Components of remote sensing

  • Energy source
  • Transmission path
  • Target
  • Sensor

Process of remote sensing

  1. The EMR (Electro-Magnetic Radiation) or solar radiation is the primary source of energy for remote sensing.
  2. Sunlight travels from the sun through the atmosphere, before it reaches the earth’s surface. In the atmosphere, the sun’s rays are not obstructed by any object.
  3. When solar radiation falls on the earth’s surface, some of its energy is absorbed. While some are transmitted through the surface, the rest is reflected. Surfaces naturally emit radiation in the form of heat. The reflected energy travels from the earth’s surface back to space.
  4. Sensors in the satellite record the reflected and emitted radiation. Each surface/object possesses a characteristic spectral signature, a unique pattern of reflecting sunlight.
  5. The energy recorded by the sensor has to be transmitted to a ground station where the data are processed into an image.
  6. The processed image is interpreted either visually by human interpreters or by computer-aided techniques called digital image processing to identify and distinguish between the different spectral signatures to get information about objects/places.
  7. Finally, we understand and apply the extracted information in mapping the area or assist in solving a particular problem.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills 86

Question 2.
Describe the
(a) Advantages of Remote Sensing
(b) Disadvantages of Remote Sensing
(c) Remote Sensing and Disaster Management
Answer:
(a) Advantages of Remote Sensing

  • It is the only practical way to obtain data from inaccessible regions, e.g. Antarctica, Amazon forest.
  • It helps in constructing cheap base maps in the absence of detailed land surveys.
  • It detects the spread of natural calamities such as flood, forest fire and volcanic eruption, so that immediate rescue operations and planning can be carried out.

(b) Disadvantages of Remote Sensing

  • It is difficult to prepare large scale maps from obtained satellite data.
  • The technique is very expensive for small areas requiring one-time analysis.

(c) Remote Sensing and Disaster Management
Remote sensing technology is highly used in disaster management to study the effects of earthquakes, tsunamis, cyclones, volcanic eruption, floods and wildfires. The preliminary data is retrieved from satellites like LANDSAT, CARTOSAT, OCEANSAT, etc. Fire and flood details can be extracted and delivered to relevant authorities within two hours of satellite image capture. E.g. major earthquakes in China and New Zealand, bushfire in Victoria and floods in Kerala. Dynamic phenomena such as floods, movement of wild animals, shoreline changes, finding lost ships and planes. Researchers use satellite imageries for these.

Samacheer Kalvi 9th Social Science Geography Solutions Chapter 7 Mapping Skills Read More »

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 1.

(i) A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7 Chinese food items. In how many ways the person can select either an Indian or Chinese food?
Solution:
Number of Indian food items = 10
Number of Chinese food items = 7
Number of ways of selecting 10 Indian food items = 10 ways
Number of ways of selecting 7 Chinese food items = 7 ways
∴ By the fundamental principle of addition, the number of ways of selecting 10 Indian food items or 7 Chinese food items is = (10 + 7) ways = 17 ways

(ii) There are 3 types of a toy cars and 2 types of toy trains available in a shop. Find the number of ways a baby can buy a toy car and a toy train?
Solution:
Given, Number of toy cars = 3
Number of toy trains = 2
∴ A baby buying a toy car from 3 can be done in 3 ways
∴ A baby buying a toy train from 2 can be done in 2 ways
∴ Buying a toy car and a toy train together can be done in 3 × 2 = 6 ways

(iii) How many two-digit numbers can be formed using 1, 2, 3, 4, 5 without repetition of digits?
Solution:
The given digits are 1, 2, 3, 4, 5 The one’s place can be filled up in 5 ways using 1, 2, 3, 4, 5 and the ten’s place can be filled up in 4 ways.
The number of two-digit numbers using the digits 1, 2, 3, 4, 5 is 4 × 5 = 20

(iv) Three persons enter into a conference hall in which there are 10 seats. In how many ways they can take their places?
Solution:
Given, Number of persons = 3 and Number of seats = 10
The first person can take his place (from 10 seats) in 10 ways
The second person can take his place (from the remaining 9 seats) in 9 ways
The third person can take his place (from the remaining 8 seats) in 8 ways
∴ The three persons together can take their places in 10 × 9 × 8 = 720 ways

(v) In how many ways 5 persons can be seated in a row?
Solution:
Number of ways of 1st person can be seated in a row = 5
Number of ways of 2nd person can be seated in a row = 4
Number of ways of 3rd person can be seated in a row = 3
Number of ways of 4th person can be seated in a row = 2
Number of ways of 5th person can be seated in a row = 1
∴ By fundamental principle of multiplication, number of ways of 5 persons can be seated in a row
= 5 × 4 × 3 × 2 × 1
= 5!
= 120

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 2.
(i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Solution:
Number of digits = 10
∴ Number of attempts made = 10 × 9 × 8 × 7 × 6 × 5 = 151200 ways

(ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use of three flags, One below the other?
Solution:
Number of flags given = 4
Number of flag needed (to show a signal) = 3
The first flag can be chosen in 4 ways (from the 4 flags)
The second flag can be chosen (from the remaining 3 flags) in 3 ways
The third flag can be chosen (from the remaining 2 flags) in 2 ways
So the first, second and the third flags together can be chosen in (to generate a signal) 4 × 3 × 2 = 24 ways
(i.e) 24 signals can be generated

Question 3.
Four children are running a race.
(i) In how many ways can the first two places be filled?
Solution:
First place can be given to any one of the 4 children and second place can be given to any one of the remaining 3 children.
Number of ways of filling the first place = 4
Number of ways of filling the second place = 3
Therefore, by the fundamental principle of multiplication total number of ways of filling the first two places is = 4 × 3 =12 ways

(ii) In how many different ways could they finish the race?
Solution:
In how many different ways could they finish the race?
The race can be finished in = 4 × 3 × 2 × 1 ways = 24 ways

Question 4.
Count the number of three-digit numbers which can be formed from the digits 2, 4, 6, 8? if.

(i) repetitions of digits is allowed
Solution:
Number of digit given = 4 (2,4, 6, 8)
So the unit place can be filled in 4 ways, 10’s place can be filled in 4 ways and 100’s place can be filled in 4 ways
∴ The unit place, 10’s place and 100’s place together can be filled (i.e) So the Number of 3 digit numbers = 4 × 4 × 4 = 64 ways

(ii) repetitions of digits is not allowed.
Solution:
Repetitions of digits is not allowed
Hundred’s Ten’s Unit
The number of ways of filling the unit place using the 4 digits 2,4,6,8 in 4 ways. A number of ways of filling the tens place using the remaining 3 digits 3 ways. The number of ways of filling the hundred’s place using the remaining 2 digits is 2 ways.
Therefore, by the fundamental principle of multiplication, the total number of 3 digit numbers without repetitions of digits is = 4 × 3 × 2 = 24 ways

Question 5.
How many three-digit numbers are there with 3 in the unit place?
(i) with repetition
Solution:
with repetition
The unit place is filled (by 3) in 1 way
The 10’s place can be filled in 10 ways
The 100’s place can be filled in 9 ways (excluding 0)
So the number of 3 digit numbers with 3 unit – place = 9 × 10 × 1 = 90

(ii) without repetition
Solution:
The unit place can be filled in only one way using the digit 3. The hundred’s place can be filled in 8 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0 and 3. The ten’s place can be filled in 8 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digit 3 and the digit placed in the hundred’s place.
Therefore, by the fundamental principle of multiplication, the total number of 3 digit numbers = 1 × 8 × 8 = 64

Question 6.
How many numbers are there between 100 and 500 with the digits 0, 1, 2, 3, 4, 5 if
(i) repetition of digits allowed
Solution:
repetition of digits allowed
The given digits are 0, 1, 2, 3, 4, 5
We have to find numbers between 100 and 500. So the 100’s place can be filled (by the numbers 1, 2, 3, 4) in 4 ways.
The 10’s place can be filled in (using 0, 1, 2, 3, 4, 5) 6 ways
and the unit-place can be filled in (using 0,1, 2, 3, 4, 5) 6 ways
But the number 100 should be excluded
So the number of numbers between 100 and 500 = 4 × 6 × 6 = 144

(ii) the repetition of digits is not allowed
Solution:

The hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and 5. Ten’s place can be filled in 5 ways using the digits 0, 1, 2, 3, 4, 5 excluding the digit placed in the hundred’s place. The unit place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding the digits placed in hundred’s place’ and ten’s place.

Therefore, by the fundamental principle of multiplication, the number of 3 digit numbers between 100 and 500 with repletion of digits using the digits 0, 1, 2, 3, 4, 5 is = 4 × 5 × 4 = 80

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 7.
How many three-digit odd numbers can be formed by using the digits 0, 1, 2, 3, 4, 5 if

(i) The repetition of digits is not allowed
Solution:
The repetition of digits is not allowed
The given digits are 0, 1, 2, 3, 4, 5. Here the odd number are 1, 3, 5.
So the unit place can be filled in 3 ways (using the 3 odd number)
After filling the unit place since 0 is a given digit be fill the 100’s place which can be
filled in
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 1
Then the 10’s place can be filled in (6 – 2) 4 ways.
So the number of 3 digit odd numbers = 3 × 4 × 4 = 48

(ii) The repetition of digits is allowed
Solution:
Since we need 3 – digit odd numbers the unit place can be filled in 3 ways using the digits 1,3 or 5. Hundred’s place can be filled in 5 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0. Ten’s place can be filled in 6 ways using the digits 0 , 1 , 2 , 3 , 4 , 5.

Therefore, by the fundamental principle of multiplication, the number of 3 – digit odd numbers formed by using the digits 0, 1 , 2 , 3 , 4, 5 with repetition of digits is = 3 × 5 × 6 = 90

Question 8.
Count the numbers between 999 and 10000 subjects to the condition that there are

(i) no restriction
Solution:
no restriction
We have to find 4 digit numbers
The 1000’s place can be filled in 9 ways (excluding zero) and the 100’s, 10’s and unit places respectively can be filled in 10, 10, 10 ways (including zero)
So the number of numbers between 999 and 10000 = 9 × 10 × 10 × 10 = 9000

(ii) no digit is repeated
Solution:
Since 0 is given as a digit we have to start filling 1000’s place.
Now 1000’s place can be filled in 9 ways (excluding 0)
Then the 100’s place can be filled in 9 ways (excluding one digit and including 0)
10’s place can be filled in (9 – 1) 8 ways and unit place can be filled in (8 – 1) 7 ways So the number of 4 digit numbers are 9 × 9 × 8 × 7 = 4536 ways

(iii) at least one of the digits is repeated
Solution:
Required number of numbers = 9000 – 4536 = 4464 numbers

Question 9.
How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if

(i) The repetition of digits are not allowed?
Solution:
The repetition of digits are not allowed.
The given digits are 0, 1, 2, 3, 4, 5. A number will be divisible by 5 if the digit in the unit place is 0 or 5
So the unit place can be filled by 0 or 5

(a) When the unit place is 0 it is filled in 1 way
And so 10’s place can be filled in 5 ways (by using 1, 2, 3, 4, 5) and 100’s place can be filled in (5 – 1) 4 ways
So the number of 3 digit numbers with unit place 0 = 1 × 5 × 4 = 20

(b) When the unit place is 5 it is filled in 1 way
Since 0 is given as a digit to fill 100’s place 0 should be excluded
So 100’s place can be filled in (excluding 0 and 5) 4 ways and 10’s place can be filled in (excluding 5 and one digit and including 0) 4 ways So the number of 3 digit numbers with unit place 5 = 1 × 4 × 4 = 16
∴ Number of 3 digit numbers ÷ by 5 = 20 + 16 = 36

(ii) The repetition of digits are allowed.
Solution:
Since the 3 – digit number is divisible by 5, the unit place can be filled in 2 ways using the digits 0 and 5. Since the repetition of digits is allowed the ten’s place is filled in 6 ways using the digits 0 , 1, 2, 3, 4,5 and the hundred’s place is filled in 5 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 .

Therefore, by the fundamental principle of multiplication, the number of 3 digit numbers formed by using the digits 0,1, 2, 3 , 4, 5 with repetition of digits is = 2 × 6 × 5 = 60

Question 10.
To travel from place A to place B, there are two different bus routes B1, B2 two different train routes T1, T2, and one air route A1. From place B to place C, there is one bus route say B1‘, two different train routes say T1‘, T2‘ and one air route A1‘. Find the number of routes of commuting from place A to place C via place B without using a similar mode of transportation.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 30
From the above diagram the number of routes from A to C
= (2 × 2 + 2 × 1) + [(2 × 1) + (2 × 1)] + [(1 × 1) + (1 × 2)]
= 4 + 2 + 2 + 2 + 1 + 2 = 13

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 11.
How many numbers are there between 1 and 1000 (both inclusive) which are divisible neither by 2 nor by 5?
Solution:
Given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Numbers which are neither divisible by 2 nor 5 should have unit place 1, 3, 7, 9.
One digit numbers:
1, 3, 7, 9 are the one-digit numbers which are neither divisible by 2 nor by 5
Therefore, the required number of one-digit numbers = 4

Two-digit numbers:
The unit place can be filled in 4 ways using the digits 1, 3, 7, 9. Ten’s place can be filled in 9 ways using all the digits excluding 0. Therefore, the required number of 2 – digit numbers = 9 × 4 = 36

Question 12.
How many strings can be formed using the letters of the word LOTUS if the word

(i) either start with L or end with S?
Solution:
either start with L or end with S?
To find the number of words starting with L
Number of letters in LOTUS = 5 when the first letter is L it can be filled in 1 way only. So the remaining 4 letters can be arranged in 4! =24 ways = n(A). When the last letter is S it can be filled in the 1 way and the remaining 4 letters can be arranged is 4! = 24 ways = n(B)
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 60
(1) (1) 3! = 6 = n(A ∩ B)
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 24 + 24 – 6 = 42
Now, neither words starts with L nor ends with S = 42

(ii) neither starts with L nor ends with S?
Solution:
Number of letters of the word LOTUS = 5.
They can be arranged in 5 ! = 120 ways
Number of words starting with L and ending with S = 42
So the number of words neither starts with L nor ends with S = 120 – 42 = 78

Question 13.
(i) Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
Solution:
Number of choices for each question = 4
Total number of questions = 6
Each question can be answered in 4 ways.
∴ The total number of ways of answering 6 questions is = 4 × 4 × 4 × 4 × 4 × 4 = 46

(ii) In how many ways 10 Pigeons can be placed in 3 different Pigeonholes?
Number of Pigeons = 10
Number of Pigeonholes = 3
Each Pigeon can occupy any of these 3 holes
∴ Total number of ways of placing 10 Pigeons
= 3 × 3 × 3 × …………….. 10 times
= 310

(iii) Find the number of ways of distributing 12 distinct prizes to 10 students?
Each price can be distributed to any one of the 10 students.
Therefore, by the rule of product, the number of ways of distributing 12 distinct prizes to 10 students are
= 10 × 10 × 10 × …………. 12 times
= 1012

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 14.
Find the value of

(i) 6!
Solution:
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(ii) 4! + 5!
Solution:
4! + 5! = (4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)
= 24 + 120 = 144

(iii) 3! – 2!
Solution:
3! – 2! = (3 × 2 × 1) – (2 × 1)
= 6 – 2 = 4

(iv) 3! × 4!
Solution:
3! × 4! = (3 × 2 × 1) × (4 × 3 × 2 × 1) = 6 × 24 = 144
12!

(v) \(\frac{12 !}{9 ! \times 3 !}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 35

(vi) \(\frac{(n+3) !}{(n+1) !}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 31

Question 15.
Evaluate \(\frac{n !}{r !(n-r) !}\) when

(i) n = 6,
r = 2
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 38
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 33

(ii) n = 10,
r = 3
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 39

(iii) For any n with r = 2
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 34

Question 16.
Find the value of n if

(i) (n + 1)! = 20(n – 1)!
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 45

(ii) \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{n}{10 !}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 66

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 Additional Questions Solved

Question 1.
If the letter of the word ‘RACHIT’ are arranged in all possible ways as listed in the dictionary, then what is the rank of the word ‘RACHIT’?
Solution:
The alphabetical order of RACHIT is A, C, H, I, R, and T
Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with 1 = 5!
and Number of words beginning with R (i.e) RACHIT = 1
∴ The rank of the word ’RACHIT’ in the dictionary = 5! + 5! + 5! + 5! + 1 = 4 × 5! + 1
= 4 × 5 . 4 . 3 . 2 . 1 + 1 = 4 × 120 + 1 = 480 + 1 = 481

Question 2.
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Solution:
Any number divisible by 5, its unit place must have 0 or 5. We have to find 4-digit numbers greater than 6000 and less than 7000.
So, the unit place can be filled with 2 ways (0 or 5) since, repetition is not allowed.
∴ Tens place can be filled with 7 ways and hundreds place can be filled with 8 ways.
But the required number is greater than 6000 and less than 7000. So, thousand places can be filled with 1 digit (i.e) 6.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 40
So, the total number of integers =1 × 8 × 7 × 2 = 112
Hence, the required number of integers = 112

Question 3.
Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
Solution:
Given that all the 5 digit numbers are greater than 7000.
So, the ways of forming 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120
Now, all the four-digit numbers greater than 7000 can be formed as follows.
Thousand places can be filled with 3 ways
Hundred places can be filled with 4 ways
Tenths place can be filled with 3 ways
Units place can be filled with 2 ways
So, the total number of 4-digits numbers = 3 × 4 × 3 × 2 = 72
∴ Total number of integers = 120 + 72 = 192
Hence, the required number of integers = 192

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

Question 4.
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 65
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 61
(c) All letters are used but the first is a vowel = 2 × 5! = 2 × 120 = 240
Hence, the required matching is
(a) ↔ (iii), (b) ↔ (i), (c) ↔ (ii)

Question 5.
Five boys and 5 girls form a line. Find the number of ways of making the seating arrangement under the following condition.
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 48
Solution:
(a) Total number of arrangement when boys and girls alternate : = (5!)2 + (5!)2
(b) No two girls sit together = 5! 6!
(c) All the girls sit together = 2! 5! 5!
(d) All the girls sit never together = 10! – 5! 6!
Hence, the required matching is (a) ↔ (iii), (b) ↔ (i), (c) ↔ (iv), (d) ↔ (ii)

Question 6.
How many automobile license plates can be made if each plate contains two different letters followed by three different digits?
Solution:
We have 26 English alphabet and 10 digits (0 to 9)
Since it is given that each plate contains 2 different letters followed by 3 different digits.
∴ Number of the arrangement of 26 letters taken 2 at a time
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 50
A three-digit number can be formed out of 10 digits = 10P3
Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 51
Total number of license plates = 650 × 720 = 468000
Hence, the required number of plates = 468000.

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Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Choose the correct or the most suitable answer.

Question 1.
If A = {(x, y) : y = ex ; x ∈ R } and B = {(x, y) : y = e-x, x ∈ R } then n(A ∩ B)
(a) Infinity
(b) 0
(c) 1
(d) 2
Solution:
(c) 1
Hint.
A∩B = (0, 1)
n(A∩B) = 1
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 1

Question 2.
IfA {(x, y) : y = sin x, x ∈ R) and 8= (x, y) : y = cos x, x ∈ R) then A∩B contains ……..
(a) no element
(b) infinitely many elements
(c) only one element
(d) cannot be determined.
Solution:
(b) infinitely many elements
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5
Question 3.
The relation R defined on a set A = {0, -1, 1, 2} by xRy if |x2 +y2| ≤ 2, then which one of the following is true?
(a) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (-1, 1), (1, 2), (1, 0)}
(b) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)
(c) Domain of R is {0, -1, 1, 2}
Solution:
(a) Range of R is {0, -1, 1}
Hint.
Given A = {0, -1, 1, 2 }
the relation R is given by x R y = |x2 + y2| ≤ 2
∴ x, y must be 0 or 1
∴ Range of R is {0, – 1 , 1 }

Question 4.
If f(x) = |x – 2| + |x + 2|, x ∈ R, then
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 2
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 3
Hint.
f(x) = |x – 2| + |x + 2|
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 4

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 5.
Let R be the set of all real numbers. Consider the following subsets of the plane R x R: S = {(x, y): y = x + 1 and 0 < x < 2} and T = {(x, y) : x – y is an integer} Then which of the following is true?
(a) T is an equivalence relation but S is not an equivalence relation.
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relations
(d) S is an equivalence relation but T is not an equivalence relation.
Solution:
(a) T is an equivalence relation but S is not an equivalence relation.
Hint.

Given R is the set of all real numbers
S = { (x, y): y = x + 1 and 0 < x < 2}
T = { (x, y): x – y is an integer} are subsets of R × R
s = { (x, y): y = x + 1 and 0 < x < 2} for x ∈ R,
x = x + 1 is not possible. ∴ (x , x) ∉ S
Hence S does not satisfy the reflexive property
∴ S is not an equivalence relation
T = {(x, y): x – y is an integer}

Reflexive:
For x ∈ R, we have x – x = 0 is an integer.
∴ (x,x)∈T forall X∈ R
Hence T satisfies reflexive property

Symmetric:
Let (x, y) ∈ T, then x – y
⇒ – (x – y) is an integer
⇒ y – x is an integer
⇒ (y, x) ∈ T
∴ T satisfies the symmetric property

Transitive:
Let (x, y), (y, z) ∈ T then x – y and y – z are integers.
⇒ x – y + y – z is an integer
⇒ x – z is an integer
⇒ (x, z) ∈ T
∴ T satisfies the transitive property
we have proved T is reflexive, symmetric, and transitive.
Thus T is an equivalence relation.

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 6.
Let A and B be subsets of the universal set N, the set of natural numbers. Then
A’ ∪ [(A ∩ B) ∪ B’] is ………
(a) A
(b) A’
(c) B
(d) N
Solution:
(d) N
Hint.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 9

Question 7.
The number of students who take both the subjects Mathematics and Chemistry is 70. This represents 10% of the enrollment in Mathematics and 14% of the enrollment in Chemistry. How many students take at least one of these two subjects?
(a) 1120
(b) 1130
(c) 1100
(d) insufficient data
Solution:
(b) 1130
Hint.
Let M denotes Mathematics students
C denotes Chemistry students
Given n(M ∩ C) = 70
10 % of the enrolement in Mathematics
Out of 100 enrolement 10 students take mathematics
∴ Number of Mathematics students n (M) = \(\frac{100}{10}\) × 70
n (M ) = 700
Number of Chemistry students n(C) = \(\frac{100}{14}\) × 70
n (C) = 500
∴ n(M ∪ C) = n(M) + n(C) – n(M ∩ C)
= 700 + 500 – 70
= 1200 – 70
= 1130
The number of students take atleast one of the subject mathematics or Chemistry = 1130

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 8.
If n[(A × B) ∩ (A × C)] = 8 and n(B ∩ C) = 2 , then n(A) is
(a) 6
(b) 4
(c) 8
(d) 16
Solution:
(b) 4

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 9.
If n(A) = 2 and n(B ∪ C) = 3, then n[(A × B) ∪ (A × C)] is …….
(a) 23
(b) 32
(c) 6
(d) 5
Solution:
(c) 6
Hint.
Given n (A) = 2 and n(B ∪ C) = 3
n[(A × B) ∪ (A × C)] = n[A × ( B ∪ C ) ]
A × (B ∪ C) = (A × B) ∪ (A × C)
= n(A) . n(B ∪ C)
= 2 × 3 = 6

Question 10.
If two sets A and B have 17 elements in common, then the number of elements common to the set A × B and B × A is
(a) 217
(b) 172
(c) 34
(d) insufficient data
Solution:
(b) 172
Hint.
n (A ∩ B) = 17
So n [(A × B) ∩ (B × A)]
= n(A ∩ B) × n(B ∩ A) = 17 × 17 = 172

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 11.
For non-empty sets A and B, if A ⊂ B then (A × B) ∩ (B × A) is equal to ……….
(a) A ∩ B
(b) A × A
(c) B × B
(d) None of these
Solution:
(b) A × A
Hint.
Given A ⊂ B, take A = { 1, 2 } and B = { 1, 2 , 3 }
A × B = {1, 2} × {1, 2, 3}
A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}
B × A = {1, 2, 3} × {1, 2}
B × A = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
(A × B) ∩ (B × A) = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)} ∩ {(1, 1), (1, 2), (2, 1),
(2, 2), (3, 1), (3, 2)}
(A × B) ∩ (B × A) = {(1, 1), (1, 2), (2, 1), (2, 2)}
A × A = {1, 2} × {1, 2}
A × A = { (1, 1), (1, 2), (2, 1), (2, 2) }
(A × B) × (B × A) = A × A

Question 12.
The number of relations on a set containing 3 elements is
(a) 9
(b) 81
(c) 512
(d) 1024
Solution:
(c) 512
Hint.
Number of relations = 2n2 = 232 = 29 = 512

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 13.
Let R be the universal relation on a set X with more than one element. Then R is
(a) Not reflexive
(b) Not symmetric
(c) Transitive
(d) None of the above
Solution:
(c) Transitive

Question 14.
Let X = {1, 2, 3, 4} and R = {(1, 1), (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1,4), (4, 1)}. Then R is ……..
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Solution:
(b) Symmetric
Hint.
x = {1, 2, 3, 4}
R = { (1, 1) , (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1, 4), (4, 1)}

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 15.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 20
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 21

Question 16.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 22
Solution:
(c) [0, 1)
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 23

Question 17.
The rule f(x) = x2 is a bijection if the domain and the co-domain are given by …..
(a) R,R
(b) R, (0, ∞)
(c) (0, ∞), R
(d) [0, ∞), [0, ∞)
Solution:
(d) [0, ∞), [0, ∞)

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 18.
The number of constant functions from a set containing m elements to a set containing n elements is
(a) mn
(b) m
(c) n
(d) m + n
Solution:
(c) n
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 25

Question 19.
The function f: [0, 2π] ➝ [-1, 1] defined by f(x) = sin x is
(a) One to one
(b) Onto
(c) Bijection
(d) Cannot be defined
Solution:
(b) Onto
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 26
So it is not one-to-one
So it is an onto function

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 20.
If the function f : [-3, 3] ➝ S defined by f(x) = x2 is onto, then S is ………
(a)[-9, 9]
(b) R
(c) [-3, 3]
(d) [0, 9]
Solution:
(d) [0, 9]

Question 21.
Let X = {1, 2, 3, 4}, Y = {a, b, c, d) and f = {(1, a), (4, b), (2, c), (3, d) (2, d)}. Then f is ………
(a) An one-to-one function
(b) An onto function
(c) A function which is not one-to-one
(d) Not a function
Solution:
(d) Not a function
Hint.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 266
Since the element 2 has two images, it is not a function

Question 22.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 28
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 29
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 30
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 45

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 23.
Let f : R ➝ R be defined by f(x) = 1 – |x|. Then the range of f is ………
(a) R
(b) (1, ∞)
(c) (-1, ∞)
(d) (-∞, 1]
Solution:
(d) (-∞, 1]
Hint.
f(x) = 1 – |x|
When x = 0, f(0) = 1 – 0 = 1
When x = – 2 , f(-2) = 1 – |- 2| = 1 – 2 = -1
When x = – 5 , f(-5) = 1 – |- 5| = 1 – 5 = -4
∴ Range of f is (- ∞, 1]

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Question 24.
The function f : R ➝ R is defined by f(x) = sin x + cos x is ……
(a) An odd function
(b) Neither an odd function nor an even function
(c) An even function
(d) Both odd function and even function
Solution:
(b) Neither an odd function nor an even function

Question 25.
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 32
(a) An odd function
(b) Neither an odd function nor an even function
(c) An even function
(d) Both odd function arid even function
Solution:
(c) An even function
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 33
Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5 34

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Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

Question 1.
Find the area of the triangle whose vertices are (0, 0), (1, 2) and (4, 3)
Solution:
Area of triangle with vertices
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 1
∴ Area of A with vertices (0, 0), (1, 2) and (4, 3) is
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 2

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

(as the area cannot be negative).

Question 2.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 3

Question 3.
Identify the singular and non-singular matrices:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 4
Solution:
(i) For a given square matrix A,
1. If |A| = 0 then it is a singular matrix.
2. If |A| ≠ 0 then it is a non singular matrix.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 5
⇒ A is a singular matrix.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 6
Which is a skew symmetric matrix
∴ |A| = 0 ⇒ A is a singular matrix.

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

Question 4.
Determine the value of a and b so that the following matrices are singular:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 7
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 8
expanding along R1
b(4 + 4) + 7 (-6 – 1) = 0 (given)
8b + 7 (-7) = 0
(i.e.,) 8b – 49 = 0 ⇒ 8b = 49 ⇒ b = 49/8

Question 5.
If cos 2θ = 0, determine \(\left[\begin{array}{ccc}{\theta} & {\cos \theta} & {\sin \theta} \\ {\cos \theta} & {\sin \theta} & {0} \\ {\sin \theta} & {0} & {\cos \theta}\end{array}\right]^{2}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 9

Question 6.
Find the value of the product; Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 10
Sol:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 11
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 12

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 Additional Problems

Question 1.
Identify the singular and non-singular matrix.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 13
Solution:
If the determinant value of a square matrix is zero it is called a singular matrix. Otherwise it is non-singular.
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 14

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

Question 2.
Show that Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 15
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 16
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 17

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Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.2

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.2

Question 1.
Verify whether the following ratios are direction cosines of some vector or not.
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 2

Question 2.
Find the direction cosines of a vectors whose direction ratios are
(i) 1, 2, 3
(ii) 3, -1, 3
(iii) 0, 0, 7
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 3

Question 3.
Find the direction cosines and direction ratios for the following vectors
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 4
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 5
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 6
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 7

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2

Question 4.
A triangle is formed by joining the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). Find the direction cosines of the medians.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 8
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 9
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 51

Question 5.
If \(\frac{1}{2}, \frac{1}{\sqrt{2}}\), a are the direction cosines of some vector, then find a.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 10

Question 6.
If (a, a + b, a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0), then find a set of values of a, b, c.
Solution:
Given (a, a + b, a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0) then find a set of values of a, b, c.
Let the given points be A (1, 0, 0) and B (0, 1, 0)
[The direction ratios of the line joining the points A (x1, y1, z1) and B (x2, y2, z2) are x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y2 – y2, z1,  z1 – z2 j
The direction ratios of the line joining the points A(1, 0, 0) and B(0, 1, 0) are
(0 – 1, 1 – 0, 0 – 0)
(- 1, 1, 0) ………. (1)
Also the direction ratios are
(1 – 0, 0 – 1, 0 – 0)
(1, -1, 0) ………. (2)
Given the direction ratios are
(a, a + b , a + b + c) ………. (3)
Comparing (1) and (3) we have
(-1, 1, 0) = (a, a + b, a + b + c)
a = -1, a + b = 1 ⇒ – 1 + b = 1 ⇒ b = 2
a + b + c = 0 ⇒ – 1 + 2 + c = 0 ⇒ c = – 1
Comparing (2) and (3) we get
a = 1 , a + b = – 1 ⇒ 1 + b = – 1 ⇒ b = – 2
a + b + c = 0 ⇒ 1 – 2 + c = 0 ⇒ c = 1
∴ The required set of values of a, b, c are
a = -1, b = 2, c = – 1 and
a = 1, b = – 2, c = 1

Question 7.
Show that the vectors \(2 \hat{i}-\hat{j}+\hat{k}, 3 \hat{i}-4 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}\) form a right angled triangle.
Sol:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 11
⇒ The given vectors form the sides of a right-angled triangle.

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2

Question 8.
Find the value of k for which the vectors \(\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}\) and \(\vec{b}=\hat{i}+\lambda \hat{j}+3 \hat{k}\) are parallel.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 12

Question 9.
Show that the following vectors are coplanar.
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 13
Solution:
Let the given three vectors be \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\). When we are able to write one vector as a linear combination of the other two vectors, then the given vectors are called coplanar vectors.
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 14
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 15
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 16
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 17
We are able to write \(\vec{a}\) as a linear combination of \(\vec{b}\) and \(\vec{c}\)
∴ The vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar

Question 10.
Show that the points whose position vectors Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 18 and Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 19 are coplanar
Solution:
Let the given points be A, B, C and D. To prove that the points A, B, C, D are coplanar, we have to prove that the vectors \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AC}}\) are coplanar
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 20
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 21
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 22
∴ we are able to write one vector as a linear combination of the other two vectors ⇒ the given vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
(i.e.,) the given points A, B, C, D are coplanar.

Question 11.
If \(\vec{a}=2 \hat{i}+3 \hat{j}-4 \hat{k}\), \(\vec{b}=3 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(\vec{c}=-3 \hat{i}+2 \hat{j}+3 \hat{k}\), find the magnitude and direction cosines of
(i) \(\vec{a}+\vec{b}+\vec{c}\)
(ii) \(3 \vec{a}-2 \vec{b}+5 \vec{c}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 23
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 24>

Question 12.
The position vectors of the vertices of a triangle are Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 25 and Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 26. Find the perimeter of the triangle
Solution:
Let A, B, C be the vertices of the triangle ABC,
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 27

Question 13.
Find the unit vector parallel to Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 28 and Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 29
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 30
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 31

Question 14.
The position vector \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) three points satisfy the relation \(2 \vec{a}-7 \vec{b}+5 \vec{c}=\overrightarrow{0}\). Are these points collinear?
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 32

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2

Question 15.
The position vectors of the points P, Q, R, S are Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 33 and Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 34 respectively. Prove that the line PQ and RS are parallel.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 35

Question 16.
Find the value or values of m for which \(m(\hat{i}+\hat{j}+\hat{k})\) is a unit vector
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 36

Question 17.
Show that points A(1, 1, 1), B(1, 2, 3), and C(2, -1, 1) are vertices of an isosceles triangle.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 37

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.2 Additional Problems

Question 1.
Show that the points whose position vectors given by
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 38
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 39
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 40
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 41

Question 2.
Find the unit vectors parallel to the sum of \(3 \hat{i}-5 \hat{j}+8 \hat{k}\) and \(-2 \hat{j}-2 \hat{k}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 42

Question 3.
The vertices of a triangle have position vectors Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 43 Prove that the triangle is equilateral.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 44
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 45

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2

Question 4.
Prove that the points Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 46 form an equilateral triangle.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 47

Question 5.
Examine whether the vectors Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 48 are coplanar
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 49
Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra - I Ex 8.2 50
⇒ We are not able to write one vector as a linear combination of the other two vectors
⇒ the given vectors are not coplanar.

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