CBSE Class 12

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 1 Relations and Functions. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Maths Chapter 1 Important Extra Questions Relations and Functions

Relations and Functions Important Extra Questions Very Short Answer Type

Question 1.
If R = {(x, y) : x + 2y = 8} is a relation in N, write the range of R.
Solution:
Range of R = {1, 2, 3}.
[∵ When x = 2, then y = 3, when x = 4, then y = 2, when x = 6, then y = 1 ]

Question 2.
Show that a one-one function :
f{1, 2, 3} → {1, 2, 3} must be onto.   (N.C.E.R.T.)
Solution:
Since ‘f’ is one-one,
∴ under ‘f’, all the three elements of {1, 2, 3} should correspond to three different elements of the co-domain {1, 2, 3}.
Hence, ‘f’ is onto.

Question 3.
What is the range of the function f(x) = \(\frac{|x-1|}{x-1}\) ? (C.B.S.E. 2010)
Solution:
When x > 1,
than f(x) = \(\frac{x-1}{x-1}\) = 1.
When x< 1,
than f(x) = \(\frac{-(x-1)}{x-1}\) = -1
Hence, R_f = {-1, 1}.

Question 4.
Show that the function f : N → N given by f(x) = 2x is one-one but not onto. (N.C.E.R.T.)
Solution:
Let x₁, x₂ ∈ N.
Now, f(x₁) = f(x₂)
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂
⇒ f is one-one.
Now, f is not onto.
∵ For 1 ∈ N, there does not exist any x ∈ N such that f(x) = 2x = 1.
Hence, f is ono-one but not onto.

Question 5.
If f : R → R is defined by f(x) = 3x + 2 find f(f(x)). C.B.S.E. 2011 (F))
Solution:
f(f(x)) = 3 f(x) + 2
= 3(3x + 2) + 2 = 9x + 8.

Question 6.
If f(x) = \(\frac{x}{x-1}\) , x ≠1 then find fof. (N.C.E.R.T)
Solution:

Question 7.
If f: R → R is defined by f(x) = (3 -x³)^1/3, find fof(x)
Solution:
fof(x) = f(f(x)) = (3-(f(x))³)^1/3
= (3 – ((3 – x³)^1/3)³)^1/3
= (3 – (3 – x³ ))^1/3= (x³)^1/3 = x.

Question 8.
Are f and q both necessarily onto, if gof is onto? (N.C.E.R.T.)
Solution:
Consider f: {1, 2, 3, 4} → {1, 2, 3, 4}
and g : {1, 2, 3,4} → {1,2.3} defined by:
f(1) = 1, f(2) = 2, f(3) = f(4) = 3
g (1) = 1, g (2) = 2, g (3) = g (4) = 3.
∴ gof = g (f(x)) {1, 2,3}, which is onto
But f is not onto.
[∵ 4 is not the image of any element]

Relations and Functions Important Extra Questions Short Answer Type

Question 1.
Let A be the set of all students of a Boys’ school. Show that the relation R in A given by:
R = {(a, b): a is sister of b} is an empty relation and the relation R’ given by :
R’ = {(a, b) : the difference between heights of a and b is less than 3 metres} is an universal relation. (N.C.E.R.T.)
Solution:
(i) Here R = {(a, b): a is sister of b}.
Since the school is a Boys’ school,
∴ no student of the school can be the sister of any student of the school.
Thus R = Φ Hence, R is an empty relation.

(ii) Here R’ = {(a,b): the difference between heights of a and b is less than 3 metres}.
Since the difference between heights of any two students of the school is to be less than 3 metres,
∴ R’ = A x A. Hence, R’ is a universal relation.

Question 2.
Let f : X → Y be a function. Define a relation R in X given by :
R = {(a,b):f(a) = f(b)}.
Examine, if R is an equivalence relation. (N.C.E.R.T.)
Solution:
For each a ∈ X, (a, a) ∈ R.
Thus R is reflexive. [∵ f (a) = f(a)]
Now (a, b) ∈ R
⇒ f(a) = f(b)
⇒ f(b) = f (a)
⇒ (b, a) ∈ R.
Thus R is symmetric.
And (a, b) ∈ R
and (b, c) ∈ R
⇒ f(a) = f(b)
and f(b) = f(c)
⇒ f(a)= f(c)
⇒ (a, c) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.

Question 3.
Let R be the relation in the set Z of integers given by:
R = {(a, b): 2 divides a – b}.
Show that the relation R is transitive. Write the equivalence class [0].   (C.B.S.E. Sample Paper 2019-20)
Solution:
Let 2 divide (a – b) and 2 divide (b – c), where a,b,c ∈ Z
⇒ 2 divides [(a – b) + (b – c)]
⇒ 2 divides (a – c).
Hence, R is transitive.
And [0] = {0, ± 2, ± 4, ± 6,…].

Question 4.
Show that the function :
f : N → N
given by f(1) = f(2) = 1 and f(x) = x -1, for every x > 2 is onto but not one-one. (N.C.E.R.T.)
Solution:
Since f(1) = f(2) = 1,
∴ f(1) = f(2), where 1 ≠ 2.
∴ ‘f’ is not one-one.
Let y ∈ N, y ≠ 1,
we can choose x as y + 1 such that f(x) = x – 1
= y + 1 – 1 = y.
Also 1 ∈ N, f(1) = 1.
Thus ‘f ’ is onto.
Hence, ‘f ’ is onto but not one-one.

Question 5.
Find gof and fog, if:
f : R → R and g : R → R are given by f (x) = cos x and g (x) = 3x². Show that gof ≠ fog. (N. C.E.R. T.)
Solution:
We have :
f(x) = cos x and g(x) = 3x².
∴ gof (x) = g (f(x)) = g (cos x)
= 3 (cos x)² = 3 cos² x
and fog (x) = f(g (x)) = f(3x²) = cos 3x².
Hence, gof ≠ fog.

Question 6.
If f(x) = \(\frac{4 x+3}{6 x-4}\) , x ≠ \(\frac{2}{3}\) find fof(x)
Solution:
We have: \(\frac{4 x+3}{6 x-4}\) …(1)
∴ fof(x) — f (f (x))

Question 7.
Let A = N x N be the set of ail ordered pairs of natural numbers and R be the relation on the set A defined by (a, b) R (c, d) iff ad = bc. Show that R is an equivalence relation.
Solution:
Given: (a, b) R (c, d) if and only if ad = bc.
(I) (a, b) R (a, b) iff ab – ba, which is true.
[∵ ab = ba ∀ a, b ∈ N]
Thus, R is reflexive.

(II) (a, b) R (c,d) ⇒ ad = bc
(c, d) R (a, b) ⇒ cb = da.
But cb = be and da = ad in N.
∴ (a, b) R (c, d) ⇒ (c, d) R (a, b).
Thus, R is symmetric.

(III) (a,b) R (c, d)
⇒ ad = bc …(1)
(c, d) R (e,f)
⇒ cf = de … (2)
Multiplying (1) and (2), (ad). (cf) – (be), (de)
⇒ af = be
⇒ (a,b) = R(e,f).
Thus, R is transitive.
Thus, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.

Question 8.
Let f: R → R be the Signum function defined as :

and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0,1] ?
Solution:
For x ∈ (0,1].

And (gof) (x) = g(f(x)) = g(1)
[∵ f(x) = 1 ∀ x > 0]
= [1] = 1
⇒ (gof) (x) = 1 ∀ x ∈ (0, 1] …(2)
From (1) and (2), (fog) and (gof) do not coincide in (0, 1].

Relations and Functions Important Extra Questions Long Answer Type 1

Question 1.
Show that the relation R on R defined as R = {(a, b):a ≤ b}, is reflexive and transitive but not symmetric.
Solution:
We have : R = {(a, b)} = a ≤ b}.
Since, a ≤ a ∀ a ∈ R,
∴ (a, a) ∈ R,
Thus, R reflexive.
Now, (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b and b ≤ c
⇒ a ≤ c
⇒ (a, c) ∈ R.
Thus, R is transitive.
But R is not symmetric
[∵ (3, 5) ∈ R but (5, 3) ∉ R as 3 ≤ 5 but 5 > 3]

Question 2.
Prove that function f : N → N, defined by f(x) = x² + x + 1 is one-one but not onto. Find inverse of f: N → S, where S is range of f.
Solution:
Let x₁, x₂ ∈ N.
Now, f(x₁) = f(x₂)

Thus, f is one-one.
Let y ∈ N, then for any x,
f(x) = y if y = x² + x + 1


Now, for y = \(\frac{3}{4}\) , x = \(-\frac{1}{2}\) ∉ N.
Thus, f is not onto.
⇒ f(x) is not invertible.
Since, x > 0, therefore, \(\frac{\sqrt{4 y-3}-1}{2}\) > 0
⇒ \(\sqrt{4 y-3}\) > 1
⇒ 4y – 3 > 1
⇒ 4y > 4
⇒ y > 1.
Redefining, f : (0, ∞) → (1, ∞) makes
f(x) = x² + x + 1 on onto function.
Thus, f (x) is bijection, hence f is invertible and f^-1 : (1, ∞) → (1,0)
f^-1(y) = \(\frac{\sqrt{4 y-3}-1}{2}\)

Question 3.
Let A = (x ∈Z : 0 ≤ x ≤ 12}.
Show that R = {(a, b) : a, b ∈ A; |a – b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2]. (C.B.S.E 2018)
Solution:
We have:
R = {(a, b): a, b ∈ A; |a – b| is divisible by 4}.
(1) Reflexive: For any a ∈ A,
∴ (a, b) ∈ R.
|a – a| = 0, which is divisible by 4.
Thus, R is reflexive.

Symmetric:
Let (a, b) ∈ R
⇒ |a – b| is divisible by 4
⇒ |b – a| is divisible by 4
Thus, R is symmetric.

Transitive: Let (a, b) ∈ R and (b, c) ∈ R
⇒ |a – b| is divisible by 4 and |b – c| is divisible by 4
⇒ |a – b| = 4λ
⇒ a – b = ±4λ ………….(1)
and |b – c| = 4µ
⇒ b – c = ± 4µ ………….(2)

Adding (1) and (2),
(a-b) + (b-c) = ±4(λ + µ)
⇒ a – c = ± 4 (λ + µ)
⇒ (a, c) ∈ R.
Thus, R is transitive.
Now, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.

(ii) Let ‘x’ be an element of A such that (x, 1) ∈ R
⇒ |x – 1| is divisible by 4
⇒ x – 1 = 0,4, 8, 12,…
⇒ x = 1, 5, 9, 13, …
Hence, the set of all elements of A which are related to 1 is {1, 5, 9}.

(iii) Let (x, 2) ∈ R.
Thus |x – 2| = 4k, where k ≤ 3.
∴ x = 2, 6, 10.
Hence, equivalence class [2] = {2, 6, 10}.

Question 4.
Prove that the function f: [0, ∞) → R given by f(x) = 9x² + 6x – 5 is not invertible. Modify the co-domain of the function f to make it invertible, and hence find f^-1.   (C.B.S.E. Sample Paper 2018-19)
Solution:
Let y∈ R.
For any x, f(x) = y if y = 9x² + 6x – 5
⇒ y = (9x² + 6x + 1) – 6
= (3x + 1)² – 6

For y = – 6 ∈ R, x = \(-\frac{1}{3}\) ∉ [0, ∞).
Thus, f(x) is not onto.
Hence, f(x) is not invertible.

We redefine,
f: [0, ∞) → [-5, ∞),
which makes f(x) = 9x² + 6x – 5 an onto function.
Now, x₁, x₂ ∈ [0, ∞) such that f(x₁) = f(x₂)
⇒ (3x₁ + 1)² = (3x₂ + 1)²
⇒[(3x₁ + 1)+ (3x₂ + 1 )][(3x₁ + 1)- (3x₂ + 1)]
⇒ [3(x₁ + x₂) + 2][3(x₁ – x₂)] = 0
⇒ x₁ = x₂
[∵ 3(x₁ + x₂) + 2 > 0]
Thus, f(x) is one-one.
∴ f(x) is bijective, hence f is invertible
and f^-1: [-5, ∞) → [0, ∞)
f^-1 (y) = \(\frac{\sqrt{y+6}-1}{3}\)

Question 5.
Check whether the relation R in the set R of real numbers, defined by :
R = {(a, b): 1 + ab > 0}, is reflexive, symmetric or transitive.   (C.B.S.E. Sample Paper 2018-19)
Solution:
R = {(a, b): 1 + ab> 0}.

Reflexive:
Now, 1 + a.a = 1 + a² > 0
⇒ (a, a) ∈ R ∀ a ∈ R.
Thus, R is reflexive.

Symmetric:
Let (a, b) ∈ R.
Then 1 + ab > 0
⇒ 1 + ba > 0
⇒ (b, a) ∈ R.
Thus, R is symmetric.

Transitive:
Take a = -8,b = -1, c = \(\frac { 1 }{ 2 }\)
Now, 1 + ab = 1 + (-8) (-1)
= 9 > 0
⇒ (a, b) ∈ R
and, 1 + bc = 1 + (-1)(\(\frac { 1 }{ 2 }\) )
= 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) > 0
⇒ (b, c) ∈ R.
But 1 + ac – 1 + (-8) (\(\frac { 1 }{ 2 }\) )
= 1 – 4 = -3 < 0
⇒ (a, c) ∉ R.
Thus, R is not transitive.
Hence, R is reflexive, symmetric but not transitive.

Question 6.
Let T be the set of all triangles in a plane with R, a relation in T given by :
R = {(T₁, T₂): T₁ is congruent to T₂}.
Show that R is an equivalence relation. (N.C.E.R.T.)
Solution:
We have:
R = {(T₁, T₂): T₁ is congruent to T₂}.
Now (T₁, T₂) ∈ R.
[ ∵ Every triangle is congruent to itself]
Thus R is reflexive.
(T₁, T₂) ∈ R
⇒ T₁ is congruent to T₂
⇒ T₂ is congruent to T₁
(T₂, T₁) ∈ R.
Thus R is symmetric.

(T₁, T₂) ∈ R and (T₂, T₃) ∈ R
⇒ T₁ is congruent to T₂ and T₂ is congruent to T₃
⇒ T₁ is congruent to T₃
(T₁, T₃) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.

Question 7.
If R₁, and R₂ are equivalence relations in a set A, show that R₁ ∩ R₂ is also an equivalence relation. (N.C.E.R.T.)
Solution:
Since R₁ and R₂ are equivalence relations, [Given]
∴ (a, a) ∈ R₁
and (a, a) ∈ R₂ ∀ a ∈ A
⇒ (a, a) ∈ R₁ ∩ R₂ ∀ ∈ G
Thus R₁ ∩ R₂ is reflexive.
Now (a, b) ∈ R₁ ∩ R₂
⇒ (a, b) ∈ R₁
and (a, b) ∈ R₂
⇒ (b, a) ∈ R₁
and (b, a) ∈ R₂
⇒ (b, a) ∈ R₁ ∩ R₂

Thus R₁ n R2 is symmetric.
And (a, b) ∈ R₁ ∩ R₂
and (b, c) ∈ R₁ ∩ R₂
⇒ (a, c) ∈ R₁
and (a, c) ∈ R₂
⇒ (a, c) ∈ R₁ ∩ R₂
Thus R₁ ∩ R₂ is transitive.
Hence, R₁ ∩ R₂ is an equivalence relation.

Question 8.
Show that f: R-{2} → R -{1} defined by:
f(x) = \(\frac{x}{x-2}\) is one-one.
Also, if g : R -{1} → R -{2} as g{x) = \(\frac{2 x}{x-1}\) find gof.
Solution:
We have: f(x) = \(\frac{x}{x-2}\) and g(x) = \(\frac{2 x}{x-1}\)

Question 9.
Let f: N → N be a function defined as :
f(x) – 9x² + 6x – 5.
Show that f: N → S, where S is the range of f, is invertible. Find the inverse of f and hence find f^-1 (43) and f^-1 (163).   (C.B.S.E. 2016)
Solution:
We have : f : N → S,
fix) = 9x² + 6x – 5.
Now f(x₁) = f(x₂)
⇒ 9x₁² + 6x₁ – 5 = 9 x₂² + 6x₂ – 5
⇒ 9(x₁² – x₂²) + 6 (x₁ – x₂) = 0
⇒ (x₁ – x₂) [9x₁ + 9x₂+ 6] = 0
⇒ x₁ = x₂ [x₁, x₂ ∈ N ]
⇒ f is one-one.
Let y ∈ S be an arbitrary number.
Now y = f(x)
⇒ y = 9x²+ 6x – 5
⇒ y = (3x + 1)² – 6
⇒ \(\sqrt{y+6}\) = 3x + 1
⇒ x = \(\frac{\sqrt{y+6}-1}{3}\) ∈ N
⇒ x = f^-1 (y).

Since f is one-one and onto
⇒ f is invertible.

Question 10.
Let f: A → B be a function defined as f(x) = \(\frac{2 x+3}{x-3}\) where A = R – {3} and B = R – {2}.
Is the function ‘f ’one-one and onto ?
Is ‘f’ invertible? If yes, then find its inverse.
Solution:
Let x₁, x₂ ∈ A = R – {3}.
Now, f(x₁) = f(x₂)
⇒ \(\frac{2 x_{1}+3}{x_{1}-3}=\frac{2 x_{2}+3}{x_{2}-3}\)
⇒ (2x₁ + 3) (x₂– 3) = (2x₂ + 3) (x₁ – 3)
⇒ 2x₁x₂ – 6x₁ + 3x₂ – 9 = 2x₁ x₂ – 6x₂ + 3x₁ – 9
⇒ – 6x₁ + 3x₂ = – 6x₂ + 3x₁
⇒ 9x₁ = 9x₂
⇒ x₁ = x₂
Thus, ‘f’ is one-one.
Let y ∈ R- {2}.
Let y = f(x₀).
Then \(\frac{2 x_{0}+3}{x_{0}-3}\) = y
⇒ 2x₀ + 3 = x₀y – 3y
⇒ x₀(y – 2) = 3(y + 1)
⇒ x₀ = \(\frac{3(y+1)}{y-2}\)
Now , y ∈ R – {2 } ⇒ \(\frac{2 x_{0}+3}{x_{0}-3}\) ∈ R – {2}

Thus, ‘f’ is onto.
Hence, ‘f’ is one-one onto and consequently ‘f’ is invertible.
Also y = \(\frac{2 x+3}{x-3}\)
xy – 3y = 2x + 3
x(y- 2) = 3(y +1)
x = \(\frac{3(y+1)}{y-2}\)
f^-1 (y) = \(\frac{3(y+1)}{y-2}\)
Hence, f^-1 (x) = \(\frac{3(x+1)}{x-2}\) for all x ∈ R – {2}.

Question 11.
Let A = R – {2} and B = R – {1}. If f: A → B is a function defined by :
f(x) = \(\frac{x-1}{x-2}\) Show that f is one one and onto.
Hence, find f^-1.
Solution:
(i) One-one : Let x₁ x₂ ∈ R – {2} such that
f(x₁) = f(x₂)
⇒ \(\frac{x_{1}-1}{x_{1}-2}=\frac{x_{2}-1}{x_{2}-2}\)
⇒ x₁x₂ – 2x₁ – x₂ + 2 = x₁x₂ – 2x₂ – x₁ + 2
⇒ x₁ = x₂
Thus,f one-one.

Onto : Let f(x) = y.
Thus \(\frac{x-1}{x-2}\) = y ⇒ x = \(\frac{2 y-1}{y-1}\)
∴ Range of f = R – {1}
= Co-domain of B
Thus f is onto.

(ii) f^-1(y) = \(\frac{2 y-1}{y-1}\)
Hence f^-1(x) = \(\frac{2 x-1}{x-1}\)

Question 12.
Show that the relation S on the set:
A = {x ∈ Z : 0 ≤ x ≤ 12} given by S = {(a, b) :a,b e Z, |a – b| is divisible by 3} is an equivalence relation.
Solution:
Reflexive :
(a, a) ∈ S
⇒ |a – a| i.e., | 0 | is divisible by 3.
Thus, S is reflexive.

Symmetric:
(a, b) ∈ S ⇒ | a – b | is divisible by 3
⇒ |b – a| is divisible by 3
⇒ (b, a) ∈ S.
Thus, S is symmetric.

Transitive :
Let (a, b) ∈ S and (b, c) ∈ S.
Thus |a – b| = 3m and |b – c| = 3n
⇒ a – b = ± 3m and b – c = ± 3 n.
∴ (a – c) = 3(± m ± n)
⇒ a – c is divisible by 3
⇒ | a – c | is divisible by 3
⇒ (a, c) ∈ S.
Thus, S is transitive.
Hence, S is an equivalence relation.

Relations and Functions Important Extra Questions Long Answer Type 2

Question 1.
Let A = (1,2,3,…, 9} and R be the relation in A x A defined by (a, b) R (c, d) if: a + d = b + c for (a, b), (c, d) in A x A.
Prove that R is an equivalence relation. Also obtain the equivalence class {(2,5)}.   (C.B.S.E. 2014)
Solution:
(i) We have : (a, b) R (c, d)
⇒ a + d = b + c on the set A = {1, 2, 3,…, 9}.
(a) (a, b) R (a, b)
⇒ a + b = b + a, which is true.
[∵ a + b = b + a ∀ a, b ∈ A]
Thus R is reflexive.

(b) (a, b) R (c, d)
⇒ a + d = b + c
(c, d) R (a, b)
⇒ c + b = d + a.
But c + b = b + c
and d + a = a + d ∀ a, b, c, d ∈ A.
∴ (a, b) R(c, d) = (c, d) R (a, b).
Thus R is symmetric.

(c) (a,b)R(c,d)
⇒ a + d = b + c ∀ a, b, c, d ∈ A ……..(1)
(c,d)R(e,f)
⇒ c + f = d + e ∀ c, d, e,f ∀ A ……………. (2)

∴ Adding (1) and (2),
(a + d) + (c +f) = (b + c) + (d + e)
⇒ a + f = b + e
⇒ (a, b) R (e, f).
Thus R is transitive.
Hence, the relation R is an equivalence relation.

(ii) {(2, 5)} = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.
[∵ 2 + 4 = 5 + 1; etc.]

Question 2.
Let N denote the set of all natural numbers and R be the relation on N x N defined by : (a, b) R(c,d) is ad(b + c) = bc(a + d).
Show that R is an equivalence relation.    (C.B.S.E. 2015)
Solution:
We have : (a, b) R (c, d)
⇒ ad (b + c) = bc (a + d) on N.
(i) {a, b) R (a, b)
⇒ ab(b + a) = ba (a + b)
⇒ ab(a + b) = ab (a + ft),
which is true.
Thus R is reflexive.

(ii) (a, b) R (c, d)
⇒ ad(b + c) = be (a + d)
⇒ bc(a + d) = ad (b + c)
⇒ cb(d + a) = da (c + b)
[∵ bc = cb and a + d = d + a;
etc. ∀ a, b, c, d ∈ N]
⇒ (cb) R (a,b).
Thus R is symmetric

(iii) Let (a , b) R (c, d) and (c, d) R ( e, f)
∴ ad (b + c) = bc (a + d)
and cf(d + e) = de(c + f)


⇒  be(a+f) = af(b+e)
⇒  af(b+e) = be(a+f)
⇒  (a, b) R (e,f).
Thus R is transitive.
Hence, R is an equivalence relation.

Question 3.
Show that the relation in the set A = {x:x ∈ W, 0 ≤ x ≤ 12} given by R = {(a, b) : (a – b) is a multiple of 4} is an equivalence relation. Also find the set of all. elements related to 2.
Solution:
We have :
R = {(a, b): (a- b) is a multiple of 4},
where a, b ∈ A = {x: x ∈ W, 0 ≤ x ≤ 12}.
(a) For any a ∈ A, we have :
(a – a) = 0, which is a multiple of 4
⇒ (a, a) ∈ R ∀ a ∈ A.
Thus R is reflexive.

(b) Let a, b ∈ A.
Now (a, b) ∈ R
⇒ (a – b) is a multiple of 4
⇒ (b – a) is a multiple of 4
⇒ (b, a) ∈ R
Thus R is symmetric.

(c) Let a, b, c ∈ A.
Now (a, b) ∈ R
and (b, c) ∈ R
⇒ (a – b) is a multiple of 4 and (b – c) is a multiple of 4
⇒ (a – b) + (b – c) is a multiple of 4
⇒ (a – c) is a multiple of 4
⇒ (a, c) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.

Question 4.
Let R be the relation defined in the set A = {1,2,3,4,5,6,7} by :
R = {(a, b) : both a and b are either odd or even}.
Show that R is an equivalence relation. Further, show that all the elements of the subset:
(1, 3, 5, 7} are related to each other and all the elements of the subset {2,4,6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2,4,6}. (N.C.E.R.T.)
Solution:
We have:
R = {{a, b): both a, b are either odd or even}.
(i) Let a ∈ A.
Both a and a are either odd or even.
∴ (a, a) ∈ R.
Thus R is reflexive.

(ii) Let (a, b) ∈ R
⇒ both a and b are either odd or even
⇒ both b and a are either odd or even
⇒ (b, a) ∈ R.
Thus R is symmetric.

(iii) Let {a, b) ∈ R and (b, c) ∈ R.
∴ Both a, b and both b, c are either odd or even
⇒ both a, c are either odd or even
⇒ (a, c) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.
Further all elements of {1,3,5,7} are related to each other.
[∵ All elements of this subset are odd] Similarly all elements of {2,4,6} are related to each other.
[∵All elements of this subset are even] But no element of {1, 3, 5, 7} is related to any element of {2, 4, 6}.
[∵Elements of {1, 3, 5, 7] are odd while elements of [2,4,6] are even]

Question 5.
Show that f : N → N given by :

is both one-one and onto.   (A.I.C.B.S.E. 2012)
Solution:
One-One.
Here we discuss the following possible cases :
(i) When x₁ is odd and x₂ is even.
Here f(x₁) = f(x₂)
⇒ x₁ + 1 = x₂ – 1
⇒ x₂ = x₁ = 2, which is impossible.

(ii) When x₁ is even and x₂ is odd.
Here f(x₁) =f(x₂)
⇒ x₁ – 1 = x₂ + 1
⇒ x₁ – x₁ = 2, which is impossible.

(iii) When x₁ and x₂ are both odd.
Here f(x₁) = f(x₂)
⇒ x₁ + 1 = x₂ + 1
⇒ x₁ = x₂
∴ ‘f’ is one-one.

(iv) When x₁ and x₂ are both even.
Here f(x₁) = f(x₂)
⇒ x₁ – 1 = x₂ – 1
⇒ x₁ = x₂
∴ ‘ f’ is one-one.

Onto. Let ‘x’ be an arbitrary natural number. When x is an odd natural number, then there exists an even natural number (x + 1) such that: f(x + 1) = (x + 1) – 1 = x.
When x is an even natural number, then there exists an odd natural number (x – 1) such that:
f(x – 1) = (x – 1) + 1 = x.
∴ Each x ∈ N has its pre-image in N.
Thus ‘f ’ is onto.
Hence, ‘f ’ is both one-one and onto.

Question 6.
Show that the function f : R → R defined by:
\(\frac{x}{x^{2}+1}\) ∀ x ∈ R is neither one-one nor onto. Also, if g : R → R is defined g(x) = 2x -1, find fog (x). (C.B.S.E. 2018)
Solution:
We have : f(x) = \(\frac{x}{x^{2}+1}\)
(i) One-one, f(x₁) = f(x₂)
⇒ \(\frac{x_{1}}{x_{1}^{2}+1}=\frac{x_{2}}{x_{2}^{2}+1}\)
⇒ x₁x₂² + x₁ = x₂x₁² + x₂
⇒  x₁x₂ (x₂ – x₁) = x₂ – x₁
x₁x₂ = 1
x₁ = \(\frac{1}{x_{2}}\)
x₁ ≠ x₂
Thus, f is not one-one.

Onto:
⇒ f(x) = y
⇒ \(\frac{x}{x^{2}+1}\) = y
⇒ x = yx² + y
⇒ x²y + y – x = 0
⇒ x can not be expressed in y.
Thus, f is not onto.
Hence, f is neither one-one nor onto.

(ii) Since g (x) = 2x – 1,
∴ fog (x) = f(g (x)) = f(2x – 1)
= \(\frac{2 x-1}{(2 x-1)^{2}+1}=\frac{2 x-1}{4 x^{2}-4 x+2}\)

Question 7.
Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x – y) is divisible by 3 is an equivalence relation.
(C.B.S.E. 2018 C)
Solution:
(x – x) = 0 is divisible by 3 for all x ∈ Z.
So, (x, x) ∈ R
:. R is reflexive.
(x – y) is divisibile by 3 implies (y – x) is divisible by 3.
So (x, y) ∈ R implies (y – x) ∈ R, x. y ∈ Z
⇒ R is symmetric.
(x – y) is divisibile by 3 and (y – z) is divisible by 3.
So (x – z) = (x – y) + (y + z) is divisible by 3.
∴ (x, z) ∈ R ⇒ R is transitive.
Hence, R is an equivalence relation.

Question 8.
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by:
f(x) = [latyex]=\left(\frac{x-2}{x-3}\right)[/latex]. Show that ‘f’ is one-one and onto and hence find f^-1. (C.B.S.E. 2012)
Solution:
Let x₁, x₂ ∈ R-{3}.
Now f(x₁) = f(x₂)
⇒ \(\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}\)
⇒ (x₁ – 2) (x₂ – 3) = (x₁ – 3) (x₂ – 2)
⇒ x₁x₂ – 3x₁ – 2x₂ + 6 = ⇒ x₁x₂ – 2x₁ – 3x₂ + 6
⇒ x₁ = x₂
Thus ‘f’ is one-one.
Let y ∈ R – {1}.
Let y = f(x₀).
Then \(\frac{x_{0}-2}{x_{0}-3}\) = y
x₀ – 2 = x₀y – 3y
x₀(y-1) = 3y – 2
x₀ = \(\frac{3 y-2}{y-1}\)
Now y ∈ R – {1}.

Thus ‘f’ is onto.
Hence, ‘f’ is one-one and onto function
⇒ ‘f’ is invertible.
Also y = \(\frac{x-2}{x-3}\)
⇒ xy — 3y = x – 2
⇒ x (y — 1) = 3y – 2
x = \(\frac{3y-2}{y-1}\)
f^-1 (y) = \(\frac{3 y-2}{y-1}\)
Hence, f^-1 (x) = \(\frac{3 x-2}{x-1}\)– for all x ∈ R-{1}.

Question 9.
Let Y = {n²: n ∈ N} ⊂ N. Consider f : N → Y as f(n) = n². Show that if is invertible. Find the inverse of ‘f.   (N.C.E.R.T.)
Solution:
Let y ∈ Y, where y is arbitrary.
Here y is of the form n², for n∈ N
⇒  n = √y.
This motivates a function :
g : Y → N, defined by g (y) = √y .

Now gof(n ) = g (f(n)) = g (n²) = √n²= n
and fog (y) =f(g (y))
= f (√y) = (√y)² = y
Thus gof = I_N
and fog = I_Y.
Hence, ‘ f ’ is invertible with f^-1 = g.

Question 10.
Let f : W → W be defined by:

Show that ‘f ’ is invertible. Find the inverse of ‘f ’. (Here ‘ W’ is the set of whole numbers) (A.I.C.B.S.E. 2015)
Solution:
We have : f : W → W defined by :

f is one-one.
When n₁ and n₂ are both odd,
then f(n₁) = f(n₂)
⇒ n₁ – 1 = n₂ – 1
⇒ n₁ = n₂
When n₁ and n₂ are both even, then
f(n₁) = f(n₂)
⇒ n₁ + 1 = n₂ + 1
⇒ n₁= n₂

∴ In both cases,
f(n₁) = f(n₂)
⇒ n₁ = n₂
When n₁is odd and n₂ is even,
then f(n₁) =  n₁ – 1, which is even
and f(n₂) = n₂ + 1, which is odd.
∴ (n₁) ≠ (n₂)
⇒ f(n₁) ≠ f(n₂)
Similarly when n₁ is even and n₂ is odd,
∴ (n₁) ≠ (n₂)
⇒ f(n₁) ≠ f(n₂)
In each case, ‘f’ is one-one.

f is onto.
When n is odd whole number, then there exists an even whole number
n – 1 ∈ W such that
f(n- 1) = (n -1) + 1 = n.
When n is even whole number, then there exists an odd whole number n + 1 ∈ W such that
f(n +1) = (n + 1) -1 = n.
Also f(1) = 0 ∈ W.
∴ each number of W has its pre-image in W.
Thus f ’ is onto.
Hence, ‘f’ is one-one onto
⇒ ‘f’ is invertible.
To obtain f^-1.
Let n₁, n₂ ∈ W
such that f(n₁) = f(n₂)
n₁ + 1 = n₂, if n₁ is even
n₁ – 1 = n₂, if n₁ is odd.

Hence f = f^-1

Important Questions for Class 12 Maths Chapter Wise Pdf with Solution 2020-2021: Here we are providing CBSE Important Extra Questions for Class 12 Maths with Solutions Pdf download in Hindi and English Medium. Students can get Class 12 Maths NCERT Solutions, Maths Class 12 Important Extra Questions and Answers designed by subject expert teachers.

CBSE Class 12th Maths Important Extra Questions and Answers Chapter Wise Pdf

Maths Important Questions Class 12 | Important Questions of Maths Class 12

1. Class 12 Maths Chapter 1 Important Questions Relations and Functions
2. Inverse Trigonometric Functions Important Questions Class 12 Maths
3. Matrices Maths Important Questions Class 12
4. Determinants Maths Class 12 Important Questions
5. Class 12 Maths Chapter 5 Important Questions Continuity and Differentiability
6. Applications of Derivatives Case Study Questions Class 12 Maths
7. Integrals Class 12 Maths Chapter Wise Important Questions
8. Applications of the Integrals Imp Questions of Maths Class 12
9. Differential Equations 12th Maths Important Questions
10. Vectors Most Important Question of Maths Class 12
11. Three Dimensional Geometry CBSE Class 12 Maths Important Questions
12. Linear Programming 12 Maths Important Question
13. Probability Class 12 Maths Extra Questions

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Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 15 Communication Systems. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 15 Important Extra Questions Communication Systems

Question 1.
flow will you classifi.’ communication systems?
Answer:
Communication systems can be classified based on the nature of source. mode of communication. type of modulation and nature of channel used.

Question 2.
What are the lips of channels used for transmission?
Answer:
i. Space communication (Broadcasting, microwave mobile etc.)
ii. Line communication (Two wire, co-axial cables, fiber optical etc.)

Question 3.
What is the length of antenna required to transmit wave of frequency 40 Hz and 40 MHz?
Answer:
The minimum length of antenna required is \(\frac {λ}{4} \)
Velocity,c = υλ
λ = \(\frac {c}{υ} \)

Question 4.
Identify the sound that can travel a longer distance – siren from a factory or horn of a car. Why?
Answer:
Siren from a factory. High intensity.

Question 5.
Mention the factors on which power of electromagnetic wave transmitted depends.
Answer:
The power of electromagnetic wave is related to the length of antenna and wavelength of the wave.
Power α \(\left(\frac{l}{\lambda}\right)^{2}\), where l is the length and λ-the wavelength.

Question 6.
Which range of wave is more reliable of intermixing – shorter or longer wavelength?
Answer:
Longer wavelength

Question 7.
is there any change in the frequency or phase due to amplitude modulation?
Answer:
No change of either frequency or phase.

Question 8.
Which physical quantity of wave is varied in AM. FM and PM?
Answer:
In AM. the physical quantity of carrier that changes is amplitude.
In FM, the physical quantity of carriers that change is frequency.
In PM, the physical quantity of carrier that changes is a phase.

Question 9.
What are the advantages and limitations of AM and FM?
answer:

Question 10.
What is ground wave propagation?
Answer:
Ground wave follows curvature of the earth and has carrier frequencies up to 2MHz. e.g. AM radio.
Ground waves progress along the surface of the earth and must be vertically polarized to prevent short circuiting the electric equipments. A wave induces currents in the ground over which it passes and thus loses some energy by absorption. This is made up by energy diffracted downwards from the upper portion of the wavefront.

There is another way also by which the ground waves get attenuated. Because of diffraction, the wavefront gradually tilts over, as shown in the figure. As the wave propagates over the earth, the tilt increases and this tilt causes greater short-circuiting of the electric component of the wave. Hence there is a reduction in the field strength. Eventually, at some distance from the antenna, the wave gets weakened and dies off. The maximum range of such a transmitter depends on its frequency and power. The ground wave propagation is effective only at VLF.

Question 11.
Ground waves are not sustained for long-range communication. Why?
Answer:
Because of damping by earth surface.

Question 12.
What is the range of frequencies used in ground wave propagation? Why?
Answer:
VLF. The attenuation of surface waves increases very rapidly with an increase in frequency.

Question 13.
How can we overcome this limitation?
Answer:
By changing to space wave communication.

Question 14.
What is the basic requirement of space wave communication?
Answer:
There should be a transmitter and receiving antenna.

Question 15.
Why is space communication also known as the line of sight communication?
Answer:
The transmitting and receiving antenna are on sight.

Question 16.
Why are repeaters needed in the line of sight communication?
Answer:
To compensate for the loss of energy during propagation.

Question 17.
What are the limitations of space wave propagation?
Answer:
This method needs repeaters and a suitable antenna length (height).

Question 18.
What is skywave communication?
Answer:
Skywave communication is otherwise called ionospheric communication. In this the electromagnetic wave of high frequency is directed towards the ionosphere which reflects the wave back to earth.

Question 19.
Which space transmission technology makes use of total internal reflection?
Answer:
ionospheric transmission.

Question 20.
is sky wave propagation possible on moon? Why?
Answer:
No. Moon has no ionosphere.

Question 21.
Can all frequencies be transmitted using sky wave propagation?
Answer:
No. Only frequencies below the critical value.

Question 22.
How does sky wave propagation depend on refractive index of atmosphere?
Answer:
The refractive index of ionosphere decreases below that of free space by the change of velocity of electrons in the ionosphere and the electromagnetic saves undergo total internal reflection.

Question 23.
Through which atmospheric layer. does the propagation take place in ground. space and sky communications?
Answer:
Ground wave – Troposphere
Space wave – Troposphere
Sky wave – Ionosphere

Question 24.
Which type of space communication has maximum range of transmission?
Answer:
Satellite

Question 25.
Compare the principle applied for each type of communication.
Answer:
Ground wave – Wireless
Space wave – Line of sight
Skywave – Total internet reflection by the ionosphere.

Question 26.
What is the range of frequency used ¡n each case?
Answer:
Ground wave – <2 MHz
Space wave -> 30 MHz
Skywave -< 10 MHz

Question 27.
Point out the limitations and uses in each case.
Answer:
Ground wave – Damping effect, wireless communication.
Space wave – Finite curvature of the earth, line of sight
Skywave – Critical frequency, long-distance coverage

Question 28.
Name the type of channel used in telephone, cable TV, and high-speed internet connections.
Answer:
Telephone – Two-wire
Cable TV – Coaxial cable
Internet – Space (satellite) .

Question 29.
Which ¡s the cheapest mode of line communication?
Answer:
Two-wire system

Question 30.
What are the merits and demerits of two wire communication?
Answer:
Signals can travel kilometres without amplification, digital and analogue signals can be sent cheap.
Attenuation of signal, interference etc.

Question 31.
Why twisted wires are preferred?
Answer:
To reduce interference of electromagnetic radiations.

Question 32.
Under which condition, does maximum power transmission occur through two wire lines?
Answer:
When the impedance of the detecting device at the receiver (load) is matched (i.e., equal) to the characteristic impedance of the two wire system.

Question 33.
What kind of cable is used to connect VC’R to TV?
Answer:
Coaxial

Question 34.
Draw the figure of coaxial cable.
Answer:

Question 35.
What is the structure of coaxial cable?
Answer:
Coaxial cables are shielded i.e., outer conductor surrounds the insulated inner wire and the outer is always grounded.

Question 36.
Which type of material  is suitable to use as spacer in coaxial cable? Why?
Answer:
Solid dielectric material, for insulation.

Question 37.
What are the merits of coaxial cables?
Answer:
These cables do not suffer from radiation problem and can he used for microwaves.

Question 38.
For establishing a communication between a transmitting and receiving station, a physical medium is used.
Answer:
(a) Name the two principal classes of communication based on the physical medium used for propagation.
(b) Construct a table showing advantages and one practical application each for the two wire, coaxial cable and optic fiber communication.
(c) in cable TV transmission usually channel in UHF band carries relatively more noise, compared to VHF band. Justify
Answer:
(a) Line communication and space communication.

(c) At higher frequency, radiation loss is high.

Question 39.
Schematic diagram for three types of satellite orbits are shown below and named as A.B.C.
Answer:

(a) Identify the polar orbit and give its approximate height from earth.
(b) Give the criteria for selecting frequency of em wave to be used in photographs from satellites.
(c) A satellite T V company attempts to use 25,000 kHz for up linking signal to a sat ellite. Say whether they have selected apt frequency. Justify.
Answer:
(a) Orbit C. Its height is about 1000 km.
(b) i. Nature of the atmosphere.
ii. Reluctance of the object.
(c) No. Because frequency below 20 MHz will undergo total internal reflection at the ionosphere.

Question 40.
The following diagrams represent some of ihe modulated signals.



Which among is following correct
a. i only
b. ii only
c. iii only
d. both i and ii
Answer:
d. both i and ii

Question 41.
in communication the concept of information is the central thing. Instead of information we deal with messages, since there is no precise definition for the word information.
(a) Name the two distinct message categories.
(b) Explain them with examples.
Answer:
(a) In electronic communication systems, we use
i. analog
ii. digital signals

(b) i. Analog signal is continuous in amplitude and time variables. e.g. Speech converted microphone signal, the ECO etc.
ii Digital – This signal is discrete in amplitude and time. Here, the analog signal is subjected to time-sampling and amplitude quantization. e.g. Digital video stream, Data files etc.

Question 42.
List the various types of communications according to
(a) nature of information
(b) mode of transmission
(c) transmission channel
(d) types of modulation
Answer:
(a) Speech. picture, fax, data transmission
(b) Analog and digital communication
(c) i. Space communication
ii. Line communication
(d) i. Sinusoidal waves – AM, FM, PM
ii. Pulsed carrier waves – PAM, PTM, PPM, PNM, PCM

Question 43.
Explain the necessity of modulation.
Answer:
The unmodified signal from the source will be usually weak to be transmitted to long distance through channel. The Long-wave signal is then suitably combined with a high frequency (short wave) wave called carrier. During combining some property of the carrier is allowed to vary in proportion to that of the signal. This process is called modulation.

Question 44.
(a) What is meant by demodulation?
(b) What is its necessity?
(c) What are the different types of demodulation?
Answer:
(a) The process of extracting the information from a modulated wave is called demodulation or detection.
(b) Demodulation is an essential process for realization at the receiving end.
(c) There are different types of detectors depending on the type of modulation. e.g. optical detector, diode detector, etc.

Question 45.
(a) What does the figure represent?
(b) What is the function of ‘C’?
(c) What is the function of ‘R’?

Answer:
(a) This is an AM demodulate (or detector)
(b) The capacitor charges to the peak voltage and then discharges through R. It serves as a capacitor filter. The diode rectifies the incoming AM signal. The rectified wave, while passing through the network, gets the RF carrier component removed thus producing the original signal.
(c) The voltage across R is the envelope of modulated wave (the signal)

Question 46.
Space wave communication is called troposphere wave propagation or LOS.
Answer:
It takes place at line of sight condition.

Question 47.
Name the four areas in which space technology finds application.
Answer:
Meteorology, climatology, oceanography, and coastal studies

Question 48.
“For long TV transmission, we need satellites “. Give reason.
Answer:
The bandwidth of picture from TV camera is about 64 Hz. Because of various factors that alternate the wave, a signal directly transmitted may not reach the destination.

Question 49.
Flow Kepler’s III law plays an important role in satellite communication?
Answer:
The stable orbit is designed by the condition that square of time period is x to the cube of mean distance of the satellite from earth.

Question 50.
(a) What do you understand by synchronous satellite?
(b) Why are such satellites used for world wide communications?
Answer:
(a) A satellite with period of revolution 24 hrs.
(b) High availability, reliability and wide coverage area.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 16 Chemistry in Everyday Life. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 16 Important Extra Questions Chemistry in Everyday Life

Chemistry in Everyday Life Important Extra Questions Very Short Answer Type

Question 1.
Name a substance that can be used as an antiseptic as well as a disinfectant. (CBSE 2019C)
Answer:
Phenol / C₆H₆₅OH

Question 2.
Differentiate between disinfectants and antiseptics. (CBSE 2012)
Answer:
Both antiseptics and disinfectants kill or prevent the growth of microorganisms. Antiseptics are applied to living tissues (cuts, wounds). But disinfectants are applied to inanimate objects only (floor, instruments).

The same substance can act as an antiseptic as well as a disinfectant by varying the concentration of the solution used. For example, 0.2% solution of phenol acts as an antiseptic and its 1% solution is a disinfectant.

Question 3.
What is the cause of a feeling of depression in human beings? Name a drug that can be useful in treating this depression. (CBSE 2012)
Answer:
The cause of depression in human beings is a low level of noradrenaline. Because of the low level of noradrenaline, the signal-sending activity of the hormones becomes low and the person suffers from depression. Phenelzine is useful in treating this depression.

Question 4.
Name one substance that can act as both:
(i) Analgesic and antipyretic. (CBSE Sample Paper 2012)
Answer:
Aspirin

(ii) Antiseptic and disinfectant. (CBSE Sample Paper 2011)
Answer:
Phenol

Question 5.
What is the tincture of iodine? What is its use? (CBSE 2019C)
Answer:
2 – 3% iodine solution of alcohol-water is called tincture of iodine. It is a powerful antiseptic and is applied on wounds.

Question 6.
(a) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate. (CBSE Delhi 2013)
Answer:
Sodium benzoate

(b) If the water contains dissolved Ca²⁺ ions, out of soaps and synthetic detergent, which will you use for cleaning clothes? (CBSE AI 2013)
Answer:
Synthetic detergent.

Question 7.
Among the following which one acts as a food preservative?
Aspartame, Aspirin, Sodium benzoate, Paracetamol
Answer:
Sodium benzoate.

Chemistry in Everyday Life Important Extra Questions Short Answer Type

Question 1.
What are food preservatives? Name two such substances. (CBSE 2012)
Answer:
Food preservatives are chemical substances that are added to food materials to prevent spoilage and to retain their nutritive value for long periods. For example, sodium benzoate, potassium metabisulphite.

Question 2.
(i) Why is bithional added to soap?
Answer:
Bithional acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(ii) Which class of drugs is used in sleeping pills? (CBSE AI 2018)
Answer:
Tranquilizers.

Question 3.
(i) What class of drug is Ranitidine?
Answer:
Antacid

(ii) Which of the following is an antiseptic?
0. 2% phenol, 1% phenol(CBSE AI 2013)
Answer:
0.2% phenol: antiseptic; 1% phenol: disinfectant

Question 4.
Give one example for each of the following:
(i) An artificial sweetener whose use is limited to cold drinks.
Answer:
Aspartame.

(ii) A non-ionic detergent. (CBSE Sample Paper 2011)
Answer:
Ester of stearic acid and polyethylene glycol
CH₃(CH₂)₁₆COO(CH₂CH₂O)_nCH₂CH₂OH

Question 5.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take their doses without consultation with the doctor. Why? (CBSE Sample Paper 2011)
Answer:
Most of the drugs taken in doses higher than recommended may cause harmful effects and act as poison leading to death. Therefore, a doctor must always be consulted before taking any medicine, who will advise the patient for proper and safe doses of the drug.

Question 6.
Why do soaps not work in hard water? (CBSE Delhi 2011)
Answer:
Hard water contains calcium and magnesium salts. Therefore, in hard water soap gets precipitated as insoluble calcium and magnesium soaps which being insoluble stick to the cloth as gummy mass and blocks the ability of soap to remove oil or grease from the cloth.

Question 7.
Explain the following term with one suitable example:
Antifertility drugs (CBSE 2010)
Answer:
Antifertility drugs. These are the chemical substances used to control the pregnancy. These are also called oral contraceptives birth control pills. The common drugs used as antifertility drugs are norethindrone, ethynylestradiol (nostril), mifepristone, ormeloxifene, etc.

Chemistry in Everyday Life Important Extra Questions Long Answer Type

Question 1.
(i) What type of drug is used in sleeping pills?
(ii) What type of detergents is used in toothpaste?
(iii) Why is the use of alitame as an artificial sweetener not recommended?
OR
Define the following terms with a suitable example in each:
(i) Broad-spectrum antibiotics
(ii) Disinfectants
(iii) Cationic detergents (CBSE Delhi 2019)
Answer:
(i) Tranquilizers
(ii) Anionic detergents
(iii) With alitame, it is difficult to control the sweetness of food to which they are added.

OR
(i) Broad-spectrum antibiotics: These are the antibiotics that are effective against several types of harmful microorganisms and are used for curing a variety of diseases, for example, chloramphenicol.

(ii) Disinfectants: The chemical substances which are used to kill microorganisms but cannot be applied on living tissues are called disinfectants, for example, phenol (1% solution).

(iii) Cationic detergents: The substances which have a long hydrocarbon chain with a positive charge on the nitrogen atom (cationic part) which is involved in cleaning action are called cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides, or bromides as anions, for example, cetyltrimethylammonium bromide.

Question 2.
What is the following substance? Give one example of it:
Antacids (CBSE2011, CBSE Delhi 2012)
Answer:
Antacids: These are the chemical substances that neutralize excess acid in the gastric juices and give relief from acid indigestion, acidity, heartburns, and gastric ulcers. Until 1970, the antacids such as sodium bicarbonate or a mixture of aluminum and magnesium hydroxide have been commonly used for the treatment of acidity.

However, excessive bicarbonate can make the stomach alkaline and trigger the production of even more acid. Nowadays, acidity is cured by drugs such as cimetidine (Tagamet), ranitidine (Zantac), omeprazole, lansoprazole, etc.

Question 3.
(i) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate
Answer:
Sodium benzoate

(ii) Why is bithional added to soap?
Answer:
Bithional acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(iii) Which class of drugs is used in sleeping pills? (CBSE Delhi 2013)
Answer:
Tranquilizers.

Question 4.
(i) What class of drug is Ranitidine?
Answer:
Antacid

(ii) If the water contains dissolved Ca²⁺ ions, out of soaps and synthetic detergent, which will you use for cleaning clothes?
Answer:
Synthetic detergent

(iii) Which of the following is an antiseptic:
0. 2% phenol, 1% phenol? (CBSE 2013)
Answer:
0.2% phenol: antiseptic; 1% phenol: disinfectant

Question 5.
(i) Give two examples of macromolecules that are chosen as drug targets.
Answer:
Carbohydrates, proteins.

(ii) What are antiseptics? Give an example.
Answer:
The chemical substances which are used to either kill or prevent the growth of micro-organisms are called antiseptics. These are not harmful to live tissues and can be safely applied on wounds, cuts, ulcers, diseased skin surfaces, etc.
For example, Dettol.

(iii) Why is the use of aspartame limited to cold foods and soft drinks? (CBSE Delhi 2014)
Answer:
Aspartame is unstable at cooking temperatures and therefore, it is used as sugar substitute for cold foods and
soft drinks.

Question 6.
(i) Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
Saccharin

(ii) What are antibiotics? Give an example.
Answer:
Antibiotics are chemical substances that are produced by micro-organisms (bacteria, fungi, moulds) and can inhibit the growth or even destroy other micro-organisms. For example, Penicillin.

(iii) Give two examples of macromolecules that are chosen as drug targets. (CBSE Delhi 2014)
Answer:
Proteins, lipids.

Question 7.
(a) Differentiate between antiseptic and disinfectant. Give one example of each.
(b) Why do we require artificial sweetening agents?
OR
Define the following terms:
(a) Tranquilizers
(b) Antacids
(c) Analgesics (CBSE2019C)
Answer:
(a) Antiseptics: Antiseptics are the chemical substances that are used to either kill or prevent the growth of micro-organisms. These are not harmful to live tissues and can be safely applied on wounds, cuts, diseased skin surfaces. For example, Dettol, Savlon, furnace, tobramycin, etc.

Disinfectants: Disinfectants are chemical substances that kill micro-organisms but cannot be applied to living tissues. In other words, they also kill micro-organisms like antiseptics but are not safe for living tissues. These are commonly applied to inanimate objects such as the floor, drainage systems, instruments, etc. Some common examples of disinfectants are phenol (1% solution), chlorine (0.2 to 0.4 ppm), etc.

(b) Artificial sweetening agents are used to reduce calorie intake. These also protect teeth from decay.
OR
(a) The chemical substances used for the treatment of stress, fatigue, mild and severe mental diseases are called tranquilizers.

Tranquilizers are neurologically active drugs that affect the message transfer mechanism from nerve to receptor. These are used to relieve or reduce mental tension, irritability, excitement, and anxiety leading to calmness. These form an essential component of sleeping pills.

(b) Acidic stomach is necessary for good health, but excessive acidity in the stomach can cause discomforts such as acid indigestion, heartburn, irritation, or pain of gastric ulcers. The chemical substances which neutralize excess acid in the gastric juices and give relief from add indigestion, acidity, heartburns, and gastric ulcers are called antacids.

(c) The chemical substances which are used to relieve pains without causing impairment of consciousness, mental confusion, incoordination or paralysis, or some other disturbances of the nervous system are called analgesics.

These are of two types:

1. Non-narcotic (non-addictive) drugs
2. Narcotic drugs

Question 8.
Explain the following terms with one suitable example for each:
(i) A sweetening agent for diabetic patients
Answer:
A sweetening agent for diabetic patients: The chemical substances which give a sweetening effect to food but do not add any calorie to our body are called artificial sweetening agents. The sweetening agent for diabetic patients is saccharin.

(ii) Enzymes
Answer:
Enzymes: Enzymes are biological catalysts produced by living cells that catalyze the biochemical reaction in living organisms. Chemically enzymes are naturally occurring simple or conjugated proteins and some enzymes are non-proteins also. These increase the rates of biochemical reactions by providing alternative paths of lower energy. For example, maltase, which catalyzes the hydrolysis of maltose to glucose.

(iii) Analgesics (CBSE AI 2011)
Answer:
Analgesics: These are medicines used to relieve pains. Aspirin and some other antipyretics are also used as analgesics. Certain narcotics (which produce sleep and unconsciousness) are also used as analgesics. For example, morphine, codeine, heroin, marijuana. These are known to be habit-forming.

Question 9.
Define and write an example for the following:
(a) Broad-spectrum antibiotics.
Answer:
Broad-spectrum antibiotics: These are the antibiotics that are effective against several types of harmful micro-organisms. Therefore, these are used for curing a variety of diseases. The common examples are tetracycline, Chloromycetin, and chloramphenicol which are effective against a variety of diseases.

Other important broad-spectrum antibiotics used are vancomycin and ofloxacin. The antibiotic dysidazirine is found to be toxic to certain strains of cancer cells.

(b) Analgesics (CBSE Sample Paper 2019)
Answer:
Analgesics: The chemical substances which are used to relieve pains without causing impairment of consciousness, mental confusion, incoordination or paralysis, or some other disturbances of the nervous system are called analgesics.

These are of two types:

1. Non-narcotic (non-addictive) drugs
2. Narcotic drugs

1. Non-narcotic drugs: The common non-addictive analgesics are aspirin and paracetamol. Aspirin (2-acetoxy benzoic acid) is the most familiar example. It inhibits the synthesis of compounds known as prostaglandins which stimulate inflammation in the tissues and cause pain. These drugs are effective in relieving skeletal pain such as that due to arthritis.

Aspirin has also been very popular because it has antipyretic (temperature lowering) properties. Now, aspirin also finds use in the prevention of heart attack because it has anti-blood-clotting action. In addition, many other potential applications of aspirin, presently under investigation, include pregnancy-related complications, viral inflammation in AIDS patients, Alzheimer’s disease, dementia, cancer, etc.

Because of the shortcomings of aspirin, other analgesics like naproxen, ibuprofen, and diclofenac sodium or potassium find use as alternatives.

2. Narcotic drugs: Certain narcotics (which produce sleep and unconsciousness) are also used as analgesics. For example, morphine and its derivatives codeine, heroin, and marijuana are used in severe pain as analgesics. These are known to be habit-forming. When used in medicinal doses, these relieve pain and produce sleep. However, in excessive (poisonous) doses these produce stupor coma, convulsions and ultimately leading to death.

These analgesics are mainly used for relief in postoperative pains, cardiac pain, and pains related to childbirth and terminal cancer.

Question 10.
(a) Why are metal hydroxides better alternatives than sodium hydrogen carbonate in antacids?
(b) Why is aspirin used in the prevention of heart attacks?
(c) Why antihistamines do not affect the secretion of acid in the stomach?
OR
Define the following terms with a suitable example of each:
(a) Tranquilizers
(b) Antibiotics
(c) Non-ionic detergents (CBSE AI 2019)
Answer:
(a) Metal hydroxides are better alternatives than sodium hydrogen carbonates because of being insoluble. These do not increase the pH above neutrality. However, excessive hydrogen carbonates can make the stomach alkaline and may trigger the production of even more acid.

(b) Aspirin is a blood thinner and has anti-blood-clotting action and therefore, used in the prevention of heart attacks.

(c) Antihistamines do not affect the secretion of acid in the stomach because anti-allergic and antacid drugs work on different receptors. The receptors present in the stomach do not react with antihistamines.
OR
(a) Tranquilizers: These are substances used to relieve mental diseases. They reduce tension and anxiety. They act on higher centers of the nervous system. These are the constituents of sleeping pills. The common examples are Derivatives of barbituric acid, Equanil, luminal, diazepam, methedrine, etc.

(b) Antibiotics: Antibiotics are chemical substances that are produced by micro-organisms (bacteria, fungi, molds) and can inhibit the growth or even destroy other micro-organisms. For example, penicillin.

(c) Non-ionic detergents: These detergents do not contain any iron in their constitution and therefore, are non-ionic like the esters of high molecular mass. However, these contain polar groups which can form hydrogen bonds with water. For example, polyethylene glycol stearate.

Question 11.
(i) Why bithional is added in soap?
(ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate?
(iii) Why soaps are biodegradable whereas detergents are non-biodegradable?
OR
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Artificial sweeteners
(iii) Analgesics (CBSE Delhi 2019)
Answer:
(i) Bithional is added to soap because it acts as an antiseptic agent and reduces the odors produced by the bacterial decomposition of organic matter on the skin.

(ii) Magnesium hydroxide is insoluble and does not increase the pH above neutrality. But excess sodium bicarbonate can make the stomach alkaline and may trigger the production of even more acid.

(iii) The synthetic detergents have hydrocarbon chains that are highly branched. Bacteria cannot degrade them easily. Therefore, detergents are non-biodegradable. On the other hand, soaps have unbranched chains which can be biodegraded more easily. Hence, soaps are biodegradable.
OR
(i) Antibiotics: Antibiotics are chemical substances that are produced by microorganisms (bacteria, fungi, and molds) and can inhibit the growth or even destroy other micro-organisms, for example, penicillin, ampicillin, Amoxycillin, etc.

(ii) Artificial sweeteners: These are the chemical compounds that give a sweetening effect to the food and enhance its odor and flavor, for example, saccharin, aspartame, alitame, etc.

(iii) Analgesics: These are neurologically active pain-killing drugs that reduce or abolish pain without causing impairment of consciousness, incoordination, mental confusion, paralysis, or some other disturbances of the nervous system, for example, aspirin, paracetamol, naproxen, ibuprofen, etc.

Question 12.
Give reasons for the following:
(i) Use of aspartame as an artificial sweetener is limited to cold foods.
Answer:
It is unstable at cooking temperature.

(ii) Metal hydroxides are better alternatives than sodium hydrogen carbonate for the treatment of acidity.
Answer:
Excessive hydrogen carbonate can make the stomach alkaline and trigger the production of even more acid. Metal hydroxides being insoluble do not increase the pH above neutrality,

(iii) Aspirin is used in the prevention of heart attacks. (CBSE Sample Paper 2018, 2019)
Answer:
Aspirin has anti-blood-clotting action.

Question 13.
(i) Why is bithional added to soap?
Answer:
To impart antiseptic properties

(ii) What is the tincture of iodine? Write its one use.
Answer:
2-3% solution of iodine in the alcohol-water mixture is the tincture of iodine. Iodine dissolved in alcohol, used as an antiseptic, and applied on wounds.

(iii) Among the following, which one acts as a food preservative?
Sodium benzoate, Aspartame (CBSE 2016)
Answer:
Sodium benzoate

Question 14.
(a) Which one of the following is a disinfectant?
0-2% solution of phenol or 1% solution of phenol
(b) What is the difference between agonists and antagonists?
(c) Write one example of each of
(i) Artificial sweetener
(ii) Antacids
OR
Define the following terms with a suitable example of each:
(a) Antiseptics
(b) Bactericidal antibiotics
(c) Cationic detergents (CBSE Al 2019)
Answer:
(a) 1% solution of phenol

(b) Agonists are substances that bind the receptor and produce a biological response.
Antagonists are the substances that bind to the receptor but inhibit its natural biological response.

(c) (i) Saccharin
(ii) Ranitidine
OR
(a) Antiseptics: The chemicals which either kill or prevent the growth of microorganisms when applied on living tissues are called antiseptics, for example, tobramycin.

(b) Bactericidal antibiotics: The chemicals which have a killing effect on microbes are called bactericidal antibiotics, for example, penicillin.

(c) Cationic detergents: The molecules which have long hydrocarbon chains with a positive charge on the nitrogen atom (cationic part) which is involved in cleansing action are called cationic detergents. These are quaternary ammonium salts of amines with acetates, chlorides, or bromides as anions. The cationic part of the molecule is involved in the cleansing action, for example, cetyltrimethylammonium bromide.

Question 15.
(i) How are synthetic detergents better than soaps?
Answer:
The detergents are better than soaps because of the following reasons:
(a) Detergents can be used for washing even in hard water. On the other hand, soaps cannot be used in hard water.
(b) Detergents can be used in acidic solutions because they are not readily decomposed in an acidic medium. On the other hand, soaps cannot be used in an acidic medium because they are decomposed into carboxylic acids in an acidic medium.
(c) Detergents have a stronger cleansing action than soap.

(ii) Can you use soaps and synthetic detergents to check the hardness of water? (CBSE 2015)
Answer:
Soaps give an insoluble precipitate of calcium and magnesium in hard water whereas detergents do not give a precipitate. Therefore, soaps but not detergents can be used to check the hardness of the water.

Question 16.
Explain the role of the allosteric site in enzyme inhibition?
Answer:
Some drugs do not bind to the enzyme’s active site. These bind to a different site of enzyme which is called the allosteric site. This bonding of inhibitor at the allosteric site changes the shape of the active site in such a way that the substrate cannot recognize it. As a result, the affinity of the substrate for the active site is reduced.

It may be noted that if the bond formed between enzyme and inhibitor is a strong covalent bond and therefore cannot be broken easily, then the enzyme gets blocked permanently. The body then degrades the enzyme-inhibitor complex and synthesizes a new enzyme.

Question 17.
What are food additives? Describe the following with suitable examples:
(i) Preservatives
(ii) Artificial sweetening agents
Answer:
Food additives are the chemicals that are added to food for their preservation and enhancing their appeal. Some common food additives are:
(a) Flavours and sweeteners
(b) Food colors (dyes)
(c) Stabilising agents
(d) Antioxidants
(e) Preservatives
(f) Nutritional supplements such as vitamins, minerals, and amino acids.

(i) Preservatives: These are the chemical substances that are added to the food materials to prevent their spoilage and to retain their nutritive value for long periods. These preservatives prevent the rancidity of food and inhibit the growth or kill the micro-organisms.

The common salt is generally added to resist the activity of micro-organisms in food. The preservation of food by adding a sufficient amount of salt to it is called salting. It is used for the preservation of raw mango, amla, beans, tamarind, fish, meat, etc. The salt prevents the water from being available for microbial growth.

Sugar syrup is also used for preserving many fruits such as apples, mango, strawberry, carrot, etc. Besides these vinegar, oils, spices, citric acid is also used as food preservatives, which are used for pickles, ketchup, jams, squashes, etc.

The growth of microbes in food material can also be prevented by adding certain chemical substances. The most common preservative used is sodium benzoate (C₆H₅COONa), salts of propanoic acid, sorbic acid, and potassium metabisulphite (source of sulfur dioxide). Certain food preservatives such as butylated hydroxyanisole (BHA) and butylated hydroxytoluene (BHT) for edible oils also act as antioxidants.

(ii) Artificial sweetening agents. These are the chemical compounds that give a sweetening effect to the food and enhance its odor and flavor. Ortho- sulphobenzimide known as saccharin is the most popular sweetening agent and has been used for many articles of food. It has a very sweet taste and is about 550 times sweeter than sucrose.

Other artificial sweeteners commercially used in food articles are aspartame (methyl ester), alitame, dulcin (urea sweetener), dihydrochalcones (DHC), sucralose, etc.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 15 Polymers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 15 Important Extra Questions Polymers

Polymers Important Extra Questions Very Short Answer Type

Question 1.
Define the term ‘homopolymerisation’ giving an example. (CBSE Delhi 2012)
Answer:
The polymers formed from one type of monomers and having same repeating units are called homopolymers. For example polyvinyl chloride (PVC), polythene.

Question 2.
Give one example of condensation polymer. (CBSE 2013)
Answer:
Nylon-6, 6

Question 3.
Which of the following is a natural polymer? (CBSE 2014)
Buna-S, Proteins, PVC
Answer:
Proteins.

Question 4.
Based on molecular forces what type of polymer is neoprene? (CBSE 2014)
Answer:
Elastomer.

Question 5.
Which of the following is a fibre?
Nylon, Neoprene, PVC (CBSE 2014)
Answer:
Nylon

Question 6.
Draw structure for the polymer used for the manufacture of non-stick utensils. (CBSE2019C)
Answer:

Question 7.
What is the primary structural feature necessary for a molecule to make it useful in a condensation polymerisation reaction? (CBSE AI 2010)
Answer:
The monomers must be bifunctional i.e. contain two functional groups.

Question 8.
Arrange the following polymers in the increasing order of tensile strength. Nylon 6, Buna-S, Polythene. (CBSE Sample Paper 2010)
Answer:
Buna – S < Polythene < Nylon 6.

Question 9.
What type of reaction occurs in the formation of Nylon 6,6 polymer? (CBSE Sample Paper 2019)
Answer:
Condensation

Question 10.

Is a homopolymer or R copolymer? (CBSE 2019C)
Answer:
Homopolymer

Question 11.
Identify the Monomers of the following polymer:

Answer:
Buta-1,3-diene, styrene.
It is obtained by the polymerisation of buta-1, 3-diene and styrene in the ratio of 3: 1 in the presence of sodium.

Polymers Important Extra Questions Short Answer Type

Question 1.
Draw the molecular structures of the monomers of
(i) PVC
Answer:
PVC : CH₂ = CH – Cl (Vinyl chloride)

(ii) Teflon (CBSE 2010)
Answer:
Teflon : F₂C = CF₂ (Tetrafluoroethene)

Question 2.
Mention two important uses of each of the following : (CBSE Delhi 2011)
(i) Bakelite
Answer:
Uses of bakelite:

• Used for making combs, fountain pens, barrels, electrical switches and handles of utensils.
• Soft bakelites are used for making glue for binding wooden laminated planks in varnishes.

(ii) Nylon 6
Answer:
Uses of Nylon 6:

• It is used for making tyre cords.
• It is used for making fabrics and ropes.

Question 3.
Draw the structure of the monomer for each of the following polymers:
(i) Nylon 6
Answer:
The monomer of Nylon-6:

(ii) Polypropene (CBSE Delhi 2012)
Answer:
The monomer of polypropene:

Question 4.
Differentiate between thermoplastic and thermosetting polymers. Give one example of each. (CBSE Delhi 2010, 2012) Answer:
Thermoplastics when heated become soft and more or less fluid. These can be moulded into any desired shape. The thermoplastics ‘ can be processed again and again.

The intermolecular forces in thermoplastics are intermediate between those of elastomers and fibres.

On the other hand, thermosetting plastics on heating become hard and insoluble masses. These cannot be moulded into the desired shape and cannot be reprocessed. The intermolecular forces in thermosetting are strong and there are cross-links that hold the molecules in place so that heating does not allow them to move freely.

The common examples are:

1. Thermoplastics: Polythene, polystyrene and polyvinyl chloride.
2. Thermosetting: Bakelite and melamine formaldehyde.

Question 5.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Bakelite
Answer:
Phenol and formaldehyde

(ii) Neoprene
Answer:
2-Chloro-1,3-butadiene(chloroprene)

Question 6.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Terylene
Answer:
Ethylene glycol, terephthalic acid

(ii) Nylon-6, 6
Answer:
Hexamethylenediamine and adipic acid.

Question 7.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Teflon
Answer:
Tetrafluoroethylene

(ii) Buna-N
Answer:
1, 3-butadiene and acrylonitrile

Polymers Important Extra Questions Long Answer Type

Question 1.
Write the structures of monomers used for getting the following polymers:
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S
OR
(i) Is it a homopolymer or copolymer? Give reason.
(ii) Write the monomers of the following polymer:

(iii) What is the role of the Sulphur in the vulcanization of rubber? (CBSE Delhi 2019)
Answer:
(i) Nylon-6,6
Hexamethylenediamine: NH₂ – (CH₂)₆ – NH₂ and Adipic acid

(ii) Glyptal

(iii) Buna – S
CH₂ = CH — CH = CH₂
1, 3 Butadiene

OR
(i) It is a homopolymer because it has only one monomer.

(iii) Sulphur makes the rubber hard, tough with greater tensile strength and non-sticky by forming sulphur crosslinks.

Question 2.
(a) Write the names of monomers of the following polymer:

Answer:
Ethylene glycol and terephthalic acid or Ethane-1,2- diol and Benzene 1, 4- dicarboxylic acid.

Terylene: It is a polymer of ethylene glycol (ethane-1,2-diol) and terephthalic acid (benzene-1, 4-dicarboxylic acid). It is obtained by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K in the presence of zinc acetate-antimony trioxide, [Zn(OOCCH₃)₂ + Sb₂O₃] catalyst. It is known as terylene or dacron.

(b) What does part 6,6 mean in the polymer Nylon-6,6?
Answer:
It represents 6 carbon atoms present in both the monomer units.

Nylon-6,6: The monomer units of nylon-6,6 are hexamethylenediamine and adipic acid. It is prepared by condensation of hexamethylene diamine with adipic acid under high pressure and high temperature.

(c) Give an example of a Biodegradable polymer. (CBSE 2019C)
Answer:
Poly-p-hydroxybutyrate-co-p-hydroxy valerate (PHBV).

It is a copolymer of 3-hydroxybutyric acid and 3-hydroxypentanoic add, in which the monomer units are joined by ester linkages.

Question 3.
Explain the following terms giving a suitable example for each :
(i) Elastomers
Answer:
Elastomers: These are polymers in which the polymer chains are held together by weak intermolecular forces. Because of the presence of weak forces, the polymer can be easily stretched. However, a few cross-linked are also introduced in the chains which impart the property of regaining the original positions after the stretching force is released. A common example is vulcanised rubber.

(ii) Condensation polymers
Answer:
Condensation polymers: A polymer formed by the condensation of two or more than two monomers is called condensation polymer. In this type, each monomer generally contains two functional groups, and condensation takes place by the loss of molecules such as H₂0, HCl, etc. Common examples are nylon-6, nylon-6,6, bakelite, terylene and alkyl resins.

(iii) Addition polymers (CBSE 2012)
Answer:
Addition polymers: A polymer formed by the direct addition of repeated monomers is called an addition polymer. In this type, the monomers are unsaturated compounds and are generally derivatives of ethene. In addition, polymers have the same empirical formula as their monomers. For example, the addition of polymers polyethene or polypropylene are obtained as:

Question 4.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
Buna-S:
1, 3-Butadiene: CH₂ = CH – CH == CH₂
Styrene: C₆H₅CH = CH₂

(ii) Neoprene
Answer:
Neoprene:
Chloroprene: CH₂ = C – CH = CH₂

(iii) Nyton-6, 6 (CBSE Delhi 2013)
Answer:
Nylon 6, 6:

• Adipic acid: HOOC (CH₂)₄ COOH
• Hexamethylenediamine: NH₂(CH₂)₆NH₂

Question 5.
Write the names and structures of the monomers of the following polymers:
(i) Bakelite
Answer:
Bakelite:

• Phenol:
  
• Formaldehyde: HCHO

(ii) Nylon-6
Answer:
Nylon-6:
Caprolactam

(iii) Polythene (CBSE Delhi 2013)
Answer:
Polythene: Ethene: CH₂ = CH₂.

Question 6.
(i) What is the role of benzoyl peroxide in the polymerisation of ethene?
(ii) Identify the monomers in the following polymer:

(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Nylon-6, 6, Polythene, Buna-S
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE 2016)
Answer:
(i) Benzoyl peroxide acts as a free radical generator. In the presence of benzoyl peroxide, phenyl free radical is formed which initiates the chain initiation step.

(iii) Buna-S < Polythene < Nylon 6, 6
OR
Mechanism
(i) Chain initiation step:

(ii) Chain propagating step:

(iii) Chain termination step:

Question 7.
Write the names and structures of the monomers of the following polymers:
(a) Neoprene
Answer:
Chloroprene CH₂ = C — CH = CH₂

(b) Bakelite
Answer:
C₆H₅OH(Phenol) and HCHO (formaldehyde)

(c) PVC (CBSE 2019C)
Answer:
Polyvinyl chloride CH₂ = CH – Cl

Question 8.
Write the structures of monomers used to obtain the following polymers:
(a) Natural rubber
(b) PVC
(c) Nylon-6,6
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE AI 2019)
Answer:
(a) Natural rubber: 2-methyl-1,3-butadiene

(b) PVC: vinyl chloride
CH₂ = CH – Cl

(c) Nylon-6,6:
HOOC-(CH₂)₄ – COOH (Adipic acid)

NH₂ — (CH₂)₆ — NH₂
(Hexamethylenedlamine)
OR
Chain Initiation Step:

Chain propagation:

Chain termination step

Question 9.
Write the structures of monomers used to obtain the following polymers:
(a) Buna-S
(b) Glyptai
(c) Nylon-6
OR
(a) Arrange the following polymers in increasing order of their intermolecular forces: Polyvinylchloride, Neoprene, Terylene
(b) Write one example of each of
(i) Natural polymer
(ii) Thermosetting polymer
(c) What is the significance of numbers 6,6 in the polymer Nylon-6,6? (CBSE Al 2019)
Answer:
(a) Buna-S
CH₂ = CH – CH = CH₂ (1,3-Butadiene) and C₆H₅CH = CH₂ (Styrene)

(b) Glyptal
HO – CH₂ — CH₂ — OH (Ethylene glycol) and

(c) Nylon-6

OR
(a) Neoprene < Polyvinyl chloride (PVC) < Terylene
(b) (i) Isoprene
(ii) Bakelite
(c) The numbers ‘6, 6’ signifies that both the monomers of nylon-6, 6 have six carbon atoms.

Question 10.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester. (CBSE AI 2013)
Answer:
Polymers that are degraded by micro-organisms within a suitable period so that the polymers and their degraded products do not cause any serious effects on the environment are called biodegradable polymers. For example,

Poly β – hydroxybutyrate – co – β – hydroxy valerate (PHBV)

Question 11.
Write the names and structures of the monomers of the following polymers:
(i) Polystyrene
Answer:
Polystyrene: Styrene:

(ii) Dacron
Answer:
Dacron:

• Ethylene glycol: HOCH₂CH₂OH
• Terephthalic acid:
  

(iii) Teflon
Answer:
Teflon:
Tetrafluoroethene: F₂C = CF₂

Question 12.
Write the names and structures of the monomers of the following polymers:
(a) Teflon
Answer:
Tetrafluoroethylene, CF₂ = CF₂

(b) Terylene
Answer:

(c) Buna-N
Answer:

Question 13.
Write the structures of monomers used to obtain the following polymers:
(a) Neoprene
(b) PHBV
(c) Bakelite
OR
(a) Arrange the following polymers in decreasing order of their intermolecular forces:
Bakelite, Polythene, Buna-S, Nylon-6,6
(b) Write the monomers of the following polymer:

(c) What is the structural difference between high-density polythene (HDP) and low-density polythene (LDP)? (CBSE AI 2019)
Answer:
(a) Neoprene:
Chloroprene

(b) PHBV

(c) Bakelite

OR
(a) Buna-S < potythene < Bakelite < Nylon-6,6

(b) HO — CH₂ — CH₂ — OH and

(c) Low-density polythene (LDP) consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore, it has low density.

High-density polythene (HDP) consists of linear chains and therefore, the molecules can be closely packed. Hence, it has a high density.

Question 14.
What are high density and low-density polythene?
Answer:
High-density polythene is obtained by heating ethene at about 333-343 K under a pressure of 6-7 atm in the presence of a catalyst such as triethylaluminium and titanium tetrachloride (known as Zeigler-Natta catalyst)

This polymer consists of linear chains and therefore the molecules can be closely packed in space. It, therefore, has a high density (0.97g/cm3) and a higher melting point (403K). It is harder, tougher and has greater tensile strength than low-density polythene.

Low-density polythene (LDP) is prepared by free radical addition and hydrogen atom abstraction. It consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore it has low density (0.92 g/cm3) and low melting point (384 K). Low-density polythene is transparent. It has moderate tensile strength and high toughness. It is chemically inert.

Question 15.
Discuss the mechanism of free radical addition polymerisation.
Answer:
A large number of unsaturated compounds such as alkenes or dienes and their derivatives are polymerised by this process. The polymerisation takes place through the generation of an initiator, which is a molecule that decomposes to form free radicals. The commonly used initiator is t-butyl peroxide.

The addition occurs as

For example, most of the commercial addition polymers are obtained from alkenes and their derivatives,

The general model of polymerisation is:

Chain initiation step:

Chain Propagating Step:

Chain terminating step

Question 16.
List some important differences between natural rubber and vulcanised rubber.
Answer:
Differences between natural rubber and vulcanised rubber:

Question 17.
(a) What is the difference between two notations: nylon-6 and nylon-6,6? Give their synthesis.
Answer:
Nylon-6 has only one compound having 6-carbon atoms while nylon-6,6 refers to a polymer obtained from 6-carbon atoms of dicarboxylic acid (adipic acid) and 6-carbon atoms of diamine (hexamethylene diamine).

Nylon-6 is synthesised from caprolactam as

Nylon-6,6 is synthesised from hexamethylenediamine and adipic acid.

(b) How is Buna-S prepared?
Answer:
Buna-S is obtained by copolymerisation of styrene and 1, 3-butadiene.

Question 18.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
1, 3-Butadiene: CH₂ = CH — CH = CH₂
Styrene: C₆H₅CH = CH₂

(ii) Buna-N
Answer:
1, 3-Butadiene: CH₂ = CH – CH = CH₂
Acrytonitnie: CH₂ = CH — CN

(iii) Dacron
Answer:

• Ethylene glycol:
  
• Terephthalic acid:
  

(iv) Neoprene.
Answer:
Chloroprene: