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Class 9 Hindi video lectures by Preeti Mam-EduGrown

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Class 9 Hindi video lectures by Preeti Mam-EduGrown

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Class 11th Chapter 1 Nature and Purpose of Business |Most Important Questions | Business Studies Class 11

Nature and Purpose of Business Class 11 Important Questions Business Studies Chapter 1

Q1. Differentiate between Economic and Non-economic activities.
Answer:
Distinction Between Economic And Non-Economic Activities:

Basis	Economic Activities	Non-Economic Activities
1. Meaning	These activities are undertaken by people to earn income to meet their material needs.	These activities are of social and religious nature and do not generate any economic gain.
2. Purpose/Motive	These activities are undertaken with economic motives, for the generation of income and wealth to earn a living.	These activities are undertaken by social or psychological motives. The basic purpose is to serve other segments of society.
3. Need Satisfaction	These satisfy the economic needs of people engaged such as food, clothing, shelter, etc.	These satisfy the social and psychological needs of people engaged.
4. Types of Examples	People engaged in economic activities include traders. Manufacturer. Teacher. Doctor. Electrician. Worker, etc.	People engaged in household activities, charitable work, social work, the welfare of the poor. etc. are said to be engaged in non¬economic activities.
5. Logic	These activities are guided by rational considerations of cost and profits.	Sentiments and emotions guide these activities.

Q2. Explain the difference between Industry, Commerce, and Trade.
Answer:
Difference between Industry, Commerce, and Trade:

Points of Difference	Industry	Commerce	Trade
1. Meaning	It relates to the production of goods and services.	It relates to the distribution of goods and services.	It relates to the actual purchase and sale of goods.
2. Capital	It requires a huge amount ’ of fixed and working capital for production	It requires limited fixed capital but huge working capital.	It requires small capital as per turnover.
3. Scope	It covers genetic, manufacturing, and constructive industries.	It covers trade and auxiliaries to trade.	Does it include? ‘Mental and external trade
4. Basis	It is the basis of modern business.	Its basis is industries and professions.	it is based upon commerce.
5. Utility	Goods are produced by transforming forms, creating form utility.	Goods are produced by a change of place or by storing. Creating place utility.	Goods are produced by transferring their possession, thus creating possession utility.
6. Place of work	Industries are established at a particular place. It may be a work-shop or factory.	Here, goods are transferred from one place to other places	The market is the place where trading activities are performed.

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Class 11th Chapter -1 Sets | NCERT Maths Solution | NCERT Solution | Edugrown

In This Post we are providing Chapter 1 Sets NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Sets Class 11 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Sets NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Class 11th Chapter 1 SETS | NCERT MATHS SOLUTION |

Chapter 1 - Sets EX 1.1 NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1

Ex 1.1 Class 11 Maths Question 1.
Which of the following are sets? Justify your answer.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter,
(ix) A collection of most dangerous animals of world.
Solution.
(i) The collection of all months of a year beginning with J is {J anuary, June, July}, which is well defined and hence it forms a set.
(ii) The collection of most talented writers of India is not well defined because opinions about ‘most talented writers’ vary from person to person and hence it does not form a set.
(iii) A team of eleven best-cricket batsmen of the world is not well defined because opinions about ‘best-cricket batsmen’ vary from person to person and hence it does not form a set.
(iv) The collection of all boys in your class is well defined and hence it forms a set.
(v) The collection of all natural numbers less than 100 is {1, 2, 3, 4,…………, 99}, which is well
defined and hence it forms a set.
(vi) A collection of novels written by the writer Munshi Prem Chand is well defined and hence it forms a set.
(vii) The collection of all even integers is {…………..,-4, -2, 0, 2, 4,……….. } which is well defined and hence it forms a set.
(viii) The collection of questions in this chapter is well defined and hence it forms a set.
(ix) A collection of most dangerous animals of the world is not well defined because opinions about ‘most dangerous animals’ vary from person to person and hence it does not form a set.

Ex 1.1 Class 11 Maths Question 2.
Let A = {1, 2, 3, 4, 5, 6). Insert the appropriate symbol ∈ or ∉ in the blank spaces:
(i) 5…A
(ii) 8…A
(iii) 0…A
(iv) 4…A
(v) 2…A
(vi) 10…A
Solution.
(i) Since 5 is the element of A. ∴ 5 ∉ A.
(ii) As 8 is not the element of A. ∴ 8 ∉ A
(iii) As 0 is not the element of A ∴ 0 ∈ A.
(iv) 4 is the element of A ∴ 4 ∈ A.
(v) 2 is the element of A ∴ 2 ∈ A.
(vi) 10 is not the element of A ∴ 10 ∉ A.

Ex 1.1 Class 11 Maths Question 3.
Write the following sets in roster form:
(i) A = {x: x is an integer and -3 < x < 7}
(ii) B = {x: x is a natural number less than 6}
(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}
(v) E = The set of all letters in the word TRIGONOMETRY
(vi) F = The set of all letters in the word
Solution.
(i) Integers lying between -3 and 7 are -2, -1, 0, 1, 2, ……….. , 6
∴ A = {-2,-1, 6}.

(ii) Natural numbers less than 6 are 1, 2, 3, 4, 5.
∴ B = 11, 2, 3, 4, 5}

(iii) Two digit natural numbers such that the sum of its digits is 8 are 17, 26, 35, 44, 53, 62, 71, 80.
∴ C= (17, 26, 35, 44, 53, 62, 71,80}

(iv) Prime number divisors of 60 are 2, 3, 5.
∴ D = (2, 3, 5}

(v) Word TRIGONOMETRY is formed by using the letters T, R, I, G, O, N, M, E, Y.
∴ E = (T, R, I, G, N, O, M, E, Y}

(vi) Word BETTER is formed by using the letters B, E, T, R
∴ F = (B, E, T, R}

Ex 1.1 Class 11 Maths Question 4.
Write the following sets in the set-builder form:
(i) {3, 6, 9, 12}
(ii) {2, 4, 8, 16, 32}
(iii) {5, 25, 125, 625}
(iv) {2,4,6,…}
(v) {1,4,9,……..,100}
Solution.
(i) Let A = (3, 6, 9, 12}
All elements of the set are natural numbers that are multiples of 3.
∴ A = (x : x = 3n, n∈N and 1 ≤ n ≤4}

(ii) Let B = (2, 4, 8, 16, 32} = (2¹, 2², 2³, 2⁴, 2⁵}
∴ B = {x : x = 2ⁿ, n ∈ N and 1 ≤ n ≤ 5}

(iii) Let C = (5, 25, 125, 625} = (5¹, 5², 5³, 5⁴}
∴ C = {x : x = 5ⁿ, n ∈ N and 1 ≤ n ≤ 4}

(iv) Let D = (2, 4, 6,……………..}
All elements of the set are even natural numbers.
∴ D = (x: x is an even natural number)

(v) Let E = {1,4,9,……….,100}
All elements of the set are perfect squares.
∴ E = {x: x = n², n ∈ N and 1 ≤ n ≤ 10}

Ex 1.1 Class 11 Maths Question 5.
List all the elements of the following sets:
(i) A = {x: x is an odd natural number}
(ii) B = {x: x is an integer, $-\frac { 1 }{ 2 }$ < x < $\frac { 9 }{ 2 }$ }
(iii) C = {x: x is an integer, x² ≤ 4}
(iv) D = {x: x is a letter in the word “LOYAL”}
(v) E = {x: x is a month of a year not having 31 days}
(vi) F = {x : x is a consonant in the English alphabet which precedes k}.
Solution.
(i) A = {x: x is an odd natural number}
∴ A = {1, 3, 5, 7,……………}

(ii) B = {x: x is an integer, $-\frac { 1 }{ 2 }$ < x < $\frac { 9 }{ 2 }$ }
∴ B = { 0, 1, 2, 3, 4}

(iii) C = {x: x is an integer, x² ≤ 4}
x² ≤ 4⇒ -2 ≤ x ≤ 2
∴ C = {-2, -1, 0, 1, 2}

(iv) D = {x: x is a letter in the word “LOYAL”}
∴ D = {L, O, Y, A}

(v) E = {x: x is a month of a year not having 31 days}
∴ E = {February, April, June, September, November}

(vi) F = {x: x is a consonant in the English alphabet which precedes k}
∴ F = {b, c, d, f, g, h, j}

Ex 1.1 Class 11 Maths Question 6.
Match each of the set on the left in the roster form with the same set on the right described in set-builder form:
i. {1,2, 3,6}   (a) {x: x is a prime number and a divisor of 6}
ii. {2,3}     (b) {x : x is an odd natural number less than 10}
iii. {M, A, T, H, E,I,C, S}   (c) {x: x is natural number and divisor of 6}

iv. {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.

Solution.
(i) → (c),
(ii) → (a),
(iii) → (d),
(iv) → (b).
The sets which are in set-builder form can be written in roster form as follows:
(a) {x : x is a prime number and a divisor of 6} = {2, 3}
(b) {x: x is an odd natural number less than 10} = {1, 3, 5, 7, 9}
(c) {x : x is natural number and divisor of 6} = {1, 2, 3, 6}
(d) {x: x is a letter of the word MATHEMATICS} = {M, A, T, H, E, I, C, S}

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Ex. 1.2 NCERT Solution

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2

Ex 1.2 Class 11 Maths Question 1.
Which of the following are examples of the null set?
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) {x: x is a natural number, x ≤ 5 and x > 7}
(iv) {y: y is a point common to any two parallel lines}
Solution.
(i) Set of odd natural numbers divisible by 2 is a null set because odd natural numbers are not divisible by 2.
(ii) Set of even prime numbers is {2} which is not a null set.
(iii) {x: x is a natural number, x < 5 and x >7} is a null set because there is no natural number which satisfies x < 5 and x > 7 simultaneously,
(iv) [y: y is a point common to any two parallel lines) is a null set because two parallel lines
do not have any common point.

Ex 1.2 Class 11 Maths Question 2.
Which of the following sets are finite or infinite?
(i) The set of months of a year
(ii) {1,2,3,…}
(iii) {1,2,3, …,99,100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
Solution.
(i) The set of months of a year is finite set because there are 12 months in a year.
(ii) {1, 2, 3, …} is an infinite set because there are infinite elements in the set.
(iii) {1, 2, 3, …, 99, 100) is a finite set because the set contains finite number of elements.
(iv) The set of positive integers greater than 100 is an infinite set because there are infinite
number of positive integers greater than 100.
(v) The set of prime numbers less than 99 is a finite set because the set contains finite number of elements.

Ex 1.2 Class 11 Maths Question 3.
State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0,0)
Solution.
(i) The set of lines which are parallel to the x-axis is an infinite set because we can draw infinite number of lines parallel to x-axis.
(ii) The set of letters in the English alphabet is a finite set because there are 26 letters in the English alphabet.
(iii) The set of numbers which are multiple of 5 is an infinite set because there are infinite multiples of 5.
(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is very large but finite.
(v) The set of circles passing through the origin (0, 0) is an infinite set because we can draw infinite number of circles passing through origin of different radii.

Ex 1.2 Class 11 Maths Question 4.
In the following, state whether A = B or not:
(i) A = {a, b, c, d};B = {d, c, b, a}
(ii) A = {4, 8, 12, 16};B = {8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}
B = {x : x is positive even integer and x≤ 10}
(iv) A = {x: x isa multiple of 10}
B = {10, 15, 20, 25, 30,…}
Solution.
(i) A = {a, b, c, d} and B = {d, c, b, a} are equal sets because order of elements does not changes a set.
∴ A = B = [a, b, c, d}.

(ii) A = {4, 8, 12, 16} and B = {8, 4, 16, 18} are not equal sets because 12 ∈ A but 12 ∉ B and 18 ∉ B but 18 ∉ A.

(iii) A = {2, 4, 6, 8,10} and B = {x: x is a positive even integer and x ≤ 10) which can be written in roster form as B = (2, 4, 6, 8, 10) are equal sets.
∴ A = B = {2, 4, 6, 8,10).

(iv) A = {x: x is a multiple of 10) can be written in roster form as A = {10, 20, 30, 40,…….. } and
B – {10, 15, 20, 25, 30, ………..} are not equal sets because 15 ∈ B but 15 ∉ A.

Ex 1.2 Class 11 Maths Question 5.
Are the following pair of sets equal ? Give reasons.
(i) A = {2, 3}, B={x: x is solution of x² + 5x + 6 = 0}
(ii) A = {x: x is a letter in the word FOLLOW}
B = {y: y is a letter in the word WOLF}
Solution.
(i) A = (2, 3} and B = {x: x is a solution of x² + 5x + 6 = 0}
Now, x² + 5x + 6 = 0 ⇒ x² + 3x + 2x + 6 = 0 ⇒ (x + 3)(x + 2) = 0 ⇒ x = -3, -2
∴ B = {-2, -3}
Hence, A and B are not equal sets.

(ii) A = {x : x is a letter in the word FOLLOW} = {F, O, L, W}
B = {y: y is a letter in the word WOLF}
= {W, O, L, F}
Hence, A = B = {F, O, L, W}.

Ex 1.2 Class 11 Maths Question 6
From the sets given below, select equal sets:
A = {2, 4, 8, 12),
B = {1, 2, 3, 4},
C = {4, 8, 12, 14},
D ={3,1,4,2},
E ={-1, 1},
F ={0, a},
G ={1, -1},
H ={0, 1}
Solution.
From the given sets, we see that sets B and D have same elements and also sets E and G have same elements.
∴ B = D = {1 ,2, 3, 4} and E = G = {-1, 1}.

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.2, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Ex. 1.3 NCERT Solution

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3

Ex 1.3 Class 11 Maths Question 1.
Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:
(i) {2, 3, 4} …{1, 2, 3, 4, 5}
(ii) {a, b, c}… {b, c, d}
(iii) {x: x is a student of Class XI of your school} … {x: x student of your school}
(iv) {x : x is a circle in the plane}… {x: x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane}… {x : x is a rectangle in the plane}
(vi) {x: x is an equilateral triangle in a plane} … {x: x is a triangle in the same plane}
(vii) {x: x is an even natural number}… {x: x is an integer}
Solution.
(i) {2, 3, 4} ⊂ {11, 2, 3, 4, 5}
(ii) [a, b, c) ⊄ {{b, c, d}
(iii) {x : x is a student of Class XI of your school} ⊂ {x : x student of your school}
(iv) {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} ⊂ {x : x is a triangle in the same plane}
(vii) {x: x is an even natural number} ⊂ {x: x is an integer}

Ex 1.3 Class 11 Maths Question 2.
Examine whether the following statements are true or false:
(i) {a, b} ⊄{b, c, a}
(ii) {a, e} ⊂ {x : x is a vowel in the English alphabet}
(iii) {1, 2, 3} ⊂ {1, 3, 5}
(iv) {a} ⊂ {a, b, c}
(v) {a} ∈ la, b, c}
(vi) {x: x is an even natural number less than 6} ⊂ {x: x is a natural number which divides 36}
Solution.

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 2

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 3

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 4

Ex 1.3 Class 11 Maths Question 3.
Let A = {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why?

Solution.

Ex 1.3 Class 11 Maths Question 4.
Write down all the subsets of the following sets
(i) {a}
(ii) {a,b}
(iii) {1,2,3}
(iv) φ
Solution.
(i) Number of elements in given set = 1
Number of subsets of given set = 2¹ = 2
∴ Subsets of given set are φ , {a}.

(ii) Number of elements in given set = 2
Number of subsets of given set = 2¹² = 4
∴ Subsets of given set are φ, {a}, {b}, {a, b}.

(iii) Number of elements in given set = 3
Number of subsets of given set = 2³ = 8
Subsets of given set are φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}.

(iv) Number of elements in given set = 0
Number of subsets of given set = 2⁰= 1
∴ Subset of given set is φ.

Ex 1.3 Class 11 Maths Question 5.
How many elements has P(A), if A = φ?
Solution.
Number of elements in set A = 0
Number of subset of set A = 2⁰ = 1
Hence, number of elements of P(A) is 1.

Ex 1.3 Class 11 Maths Question 6.
Write the following as intervals:
(i) {x: x ∈ R, -4 < x ≤ 6}
(ii) {x: x ∈ R, -12 < x < -10}
(iii) {x: x ∈ R, 0 ≤ x < 7}
(iv) {x: x ∈ R, 3 ≤ x ≤ 4}
Solution.
(i)Let A = {x: x ∈ R, -4 < x ≤ 6}
It can be written in the form of interval as (-4, 6)
(ii) Let A= {x: x ∈ R, -12 < x < -10}
It can be written in the form of interval as (-12, -10)
(iii) Let A = {x: x ∈ R, 0 ≤ x < 7}
It can be written in the form of interval as (0, 7).
(iv) Let A = {x: x ∈ R, 3 ≤ x ≤ 4}
It can be written in the form of interval as (3,4).

Ex 1.3 Class 11 Maths Question 7.
Write the following intervals in set-builder form:
(i) (-3,0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [-23, 5)
Solution.
(i) The interval (-3, 0) can be written in set-builder form as {x : x ∈ R,-3 < x < 0}.
(ii) The interval [6, 12] can be written in set-builder form as {x : x ∈ R, 6 ≤ x ≤ 12}.
(iii) The interval (6, 12] can be written in set-builder form as {x : x ∈ R, 6 < x ≤ 12}
(iv) The interval [-23,5) can be written in set-builder form as {x : x ∈ R, -23 ≤ x < 5}

Ex 1.3 Class 11 Maths Question 8.
What universal set(s) would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Solution.
(i) Right triangle is a type of triangle. So the set of triangles contain all types of triangles.
∴ U = {x : x is a triangle in a plane}

(ii) Isosceles triangle is a type of triangle. So the set of triangles contain all types of triangles.
∴ U = }x : x is a triangle in a plane}

Ex 1.3 Class 11 Maths Question 9.
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set(s) for all the three sets A, B and C
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) φ
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}
Solution.

(i) {0, 1, 2, 3, 4, 5, 6} is not a universal set for A, B, C because 8 ∈ C but 8 is not a member of {0, 1, 2, 3, 4, 5, 6}.
(ii) φ is a set which contains no element. So it is not a universal set for A, B, C.
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a universal set for A, B, C because all members of A, B, C are present in {0,1 , 2, 3, 4, 5, 6, 7, 8, 9, 10).
(iv) (1, 2, 3, 4, 5, 6, 7, 8) is not a universal set for A, B, C because 0 ∈ C but 0 is not a member of {1, 2, 3, 4, 5, 6, 7, 8)

We hope the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.3, drop a comment below and we will get back to you at the earliest.

Chapter-1 Sets Ex. 1.4 NCERT Solution

Ex 1.4 Class 11 Maths Question 1.
Find the union of each of the following pairs of sets:
(i) X = {1 ,3, 5}, Y= {1, 2, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} B = (x:x is a natural number and 6 <x< 10}
(v) A = {1, 2, 3}, B = φ
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 1

Ex 1.4 Class 11 Maths Question 2.
Let A = {a, b}, B = {a, b, c}. Is A ⊂ B ? What is A ∪B?
Solution.
Here A = {a, b} and B = {a, b, c}. All elements of set A are present in set B.
∴ A ⊂ B. Now, A ∪ B = {a, b, c) = B.

Ex 1.4 Class 11 Maths Question 3.
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution.
Here A and B are two sets such that A ⊂ B.
Take A = {1, 2} and B = {1, 2, 3}.
A ∪ B = {1, 2, 3) = B.

Ex 1.4 Class 11 Maths Question 4.
If A = {11, 2, 3, 4}, B = {3, 4, 5, 6}, C={5, 6, 7, 8} and D = {7, 8, 9, 10}; find
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ O
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Solution.
Here A = {11, 2, 3, 4}, B = {3, 4, 5, 6}, C={5, 6, 7, 8} and D = {7, 8, 9, 10}
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 2

Ex 1.4 Class 11 Maths Question 5.
Find the intersection of each pair of sets of .
(i) X = {1 ,3, 5}, Y= {1, 2, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6} B = (x:x is a natural number and 6 <x< 10}
(v) A = {1, 2, 3}, B = φ
Solution.
(i) Here X = {1, 3, 5} and Y = {1, 2, 3}
∴ X ∩ Y= {1,3}

(ii) Here A = {a, e, i, o, u} and B = {a, b, c}
∴ A ∩ B = {a}

(iii) Here A = {x: x is a natural number and multiple of 3} = {3, 6, 9,12,….} and B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5} ∴ A ∩ B = {3}

(iv) Here A = {x: x is a natural number and 1 < x < 6} ={2, 3, 4, 5, 6} and B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9} ∴ A ∩ B = φ

(v) Here A = {1, 2, 3) and B = φ
∴ A ∩ B = φ

Ex 1.4 Class 11 Maths Question 6.
If A = (3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find
(i) A ∩ B
(ii) B ∩ C
(iii) A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∪ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Solution.
Here A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 3

Ex 1.4 Class 11 Maths Question 7.
If A = {x: x is a natural number), B = {x: x is an even natural number}, C={x : x is an odd natural number} and D = {x: x is a prime number}, find
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Solution.
Here A = {x: x is a natural number}
= (1, 2, 3, 4, 5, …….}
B = {x: x is an even natural number}
= 12, 4, 6,………}
C = {x: x is an odd natural number}
= {1, 3, 5, 7,………}
and D = {x: x is a prime number}
= {2, 3, 5, 7,….}

(i) A ∩ B = {x: x is a natural number} ∩ {x: x is an even natural number}
= {x: x is an even natural number} = B.

(ii) A ∩ C = {x: x is a natural number} ∩ {x: x is an odd natural number}
= {x: x is an odd natural number} = C.

(iii) A ∩ D = {x: x is a natural number} ∩ {x: x is a prime number}
= {x: x is a prime number} = D.

(iv) B ∩ C = {x: x is an even natural number} ∩{x: x is an odd natural number} = φ .

(v) B ∩ D = [x: x is an even natural number} ∩ {x: x is a prime number} = {2}.

(vi) C ∩ D = {x: x is an odd natural number} ∩ {x: x is a prime number} = {x: x is an odd prime number}.

Ex 1.4 Class 11 Maths Question 8.
Which of the following pairs of sets are disjoint?
(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}
(ii) {a, e, i, o, u] and {c, d, e, f}
(iii) {x: x is an even integer} and {x: x is an odd integer}
Solution.
(i) Let A = {1,2,3,4}
and B = {x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
∴ A ∩ B = {1,2,3,4} n {4,5, 6} = {4}
Hence A and B are not disjoint sets.

(ii) Let A = {a, e, i, o, u} and B = {c, d, e, f}
∴ A ∩ B = {e}
Hence A and B are not disjoint sets.

(iii) Let A = {x : x is an even integer} and B = {x: x is an odd integer}
∴ A ∩ B = φ. Hence A and B are disjoint sets.

Ex 1.4 Class 11 Maths Question 9.
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}; find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii)D – C
Solution.
Here A = {3, 6, 9, 12, 15, 18, 21},
B = {4, 8, 12, 16, 20},
C ={2, 4, 6, 8, 10, 12, 14, 16},
D = {5, 10, 15, 20}
(i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8,12,16, 20} = {3, 6, 9,15,18, 21}
(ii) A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16} = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5,10,15, 20} = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21} = {4, 8,16, 20}
(v) C – A = {2,4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21} = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21} = {5, 10, 20}
(vii) B – C={4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {20}
(viii) B – D = {4, 8, 12, 16, 20} – {5, 10, 15, 20} = {4, 8, 12, 16}
(ix) C – B = {2,4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20} = {2, 6, 10, 14}
(x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20} = {5, 10, 15}
(xi) C – D = {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20} = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C={5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {5, 15, 20}

Ex 1.4 Class 11 Maths Question 10.
If X= {a, b, c, d} and Y={f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
Solution.
Here X = {a, b, c, d} and Y = {f, b, d, g}
(i) X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}
(ii) Y – X = {f, b, d, g} – {a, b, c, d} = {f, g}
(iii) X ∩ Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}

Ex 1.4 Class 11 Maths Question 11.
If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Solution.
We know that set of real numbers contain rational and irrational numbers. So R – Q = set of irrational numbers.

Ex 1.4 Class 11 Maths Question 12.
State whether each of the following statement is true or false. Justify your answer.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 5

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Chapter-1 Sets Ex. 1.5 NCERT Solution

Ex 1.5 Class 11 Maths Question 1.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = { 1, 2, 3, 4}, B = (2,4,6,8} and C = {3,4,5,6}. Find
(i) A’
(ii) B’
(iii) (A ∪ C)’
(iv) (A ∪B)’
(v) (A’)’
(vi) (B – C)’
Solution.
Here U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C ={3, 4, 5, 6}
(i) A’=U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}

(ii) B’=U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}

(iii) A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
= (1, 2, 3, 4, 5, 6}
(A∪C)’=U-(A∪C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}
= {7, 8, 9}

(iv) A ∪ B = {1, 2, 3,4} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 6, 8}
(A∪B)’ = U – (A∪B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
= {5, 7, 9}

(v) We know that A’ = {5, 6, 7, 8, 9}
(A’)’ =U – A’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
= {1, 2, 3, 4}

(vi) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}
(B-C)’=U – (B-C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
= {1, 3, 4, 5, 6, 7, 9}.

Ex 1.5 Class 11 Maths Question 2.
If U = {a,b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
Solution.
(i) A’ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}
= {d, e,f, g, h}

(ii) B’ = U – B = {a, b, c, d, e,f, g, h} – {d, e, f, g}
= {a, b, c, h}

(iii) C’ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(iv) D’ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}
= {b, c, d, e}.

Ex 1.5 Class 11 Maths Question 3.
Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x: x is an even natural number}
(ii) {x: x is an odd natural number}
(iii) {x: x is a positive multiple of 3}
(iv) {x: x is a prime number}
(v) {x: x is a natural number divisible by 3 and 5}
(vi) {x: x is a perfect square}
(vii) {x: x is a perfect cube}
(viii) {x: x + 5 = 8}
(ix) (x: 2x + 5 = 9)
(x) {x: x ≥ 7}
(xi) {x: x ∈ W and 2x + 1 > 10}
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 1

Ex 1.5 Class 11 Maths Question 4.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) (A ∪ B)’ = A’∩B’
(ii) (A ∩ B)’ = A’∪B’
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 4

Ex 1.5 Class 11 Maths Question 5.
Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)’
(ii) A’∩B’
(iii) (A ∩ B)’
(iv) A’ ∪ B’
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 6

Ex 1.5 Class 11 Maths Question 6.
Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?
Solution.
Here U = {x : x is a triangle}
A = {x: x is a triangle and has at least one angle different from 60°}
∴ A’ = U – A = {x : x is a triangle} – {x : x is a triangle and has atleast one angle different from 60°}
= {x : x is a triangle and has all angles equal to 60°)
= Set of all equilateral triangles.

Ex 1.5 Class 11 Maths Question 7.
Fill in the blanks to make each of the following a true statement:
(i) A ∪ A’ = …….
(ii) φ’ ∩ A = .…….
(iii) A ∩ A’ = …….
(iv) U’ ∩ A = .…….
Solution.
(i) A ∪ A’= U
(ii) φ’ ∩ A = U ∩ A = A
(iii) A ∩ A’ = φ
(iv) U’ ∩ A = φ ∩ A = φ

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Chapter-1 Sets Ex. 1.6 NCERT Solution

Ex 1.6 Class 11 Maths Question 1.
If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n (X ∪ Y) = 38, find n(X ∩ Y).
Solution.
Here n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38
We know that
n(X ∪ Y) = n(X) + n(Y) -n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
∴ n(X ∩ Y) = 40 – 38 = 2.

Ex 1.6 Class 11 Maths Question 2.
If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?
Solution.
Here n(X ∪ Y) = 18. n(X) = 8 and n(Y) = 15
We know that
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 18 = 8 +15 – n(X ∩ Y)
∴ n(X ∩ Y) = 23-18 = 5.

Ex 1.6 Class 11 Maths Question 3.
In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Solution.
Let H be the set of people speaking Hindi and E be the set of people speaking English.
∴ n(H) = 250, n(E) = 200 and n(H ∪ E) = 400,
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
400 = 250 + 200 – n(H ∩ E)
∴ n(H ∩ E) = 450 – 400 = 50.

Ex 1.6 Class 11 Maths Question 4.
If 5 and Tare two sets such that 5 has 21 elements, T has 32 elements, and S ∩T has 11 elements, how many elements does S ∪ T have?
Solution.
Here n(S) = 21, n(T) = 32 and n(S ∩T) = 11
We know that
n(S ∪ T) = n(S) + n(T) – n(S ∩ T) n(S ∪ T)
= 21 + 32 – 11 = 42.

Ex 1.6 Class 11 Maths Question 5.
If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements, and X ∩ Y has 10 elements, how many elements does X have?
Solution.
Here n{X) = 40, n(X ∪ Y) = 60 and n(X ∩ Y) = 10
We know that
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
∴ n(Y) = 60 – 30 = 30.

Ex 1.6 Class 11 Maths Question 6.
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution.
Let C be the set of persons who like coffee and T be the set of persons who like tea.
∴ n(C) = 37, n(T) = 52 and n(C ∪ T) = 70
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 70 = 37 + 52 – n(C ∩ T)
∴ n(C ∩ T) = 89 – 70 = 19.

Ex 1.6 Class 11 Maths Question 7.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution.
Let C be the set of people who like cricket and T be the set of people who like tennis. Here n(Q) = 40, n(C ∩ T) = 10 and n(C ∪ T) = 65 .
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n(T) -10
⇒ n{T} = 65 – 30 = 35
∴ Number of people who like tennis = 35 j Now number of people who like tennis only and not cricket
=n(T – C) = n(T) – n(C ∩ T) = 35 – 10 = 25.

Ex 1.6 Class 11 Maths Question 8.
In a committee, 50 people speak French, 20 f speak Spanish and 10 speak both Spanish and
French. How many speak at least one of these two languages?
Solution.
Let F be the set of people who speak French and S be the set of people who speak Spanish.
Here n(F) = 50, n(S) = 20 and n(F ∩ S) = 10
We know that
n(F ∪ S) = n(F) + n(S) – n(F ∩ S)
n(F ∪ S) = 50 + 20 -10 = 60
∴ Number of people who speak at least one of these two languages = 60

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Chapter-1 Sets Mislleneous NCERT Solution

Miscellaneous Exercise Class 11 Maths Question 1.
Decide, among the following sets, which sets are subsets of one and another:
A = {x : x ∈R and x satisfy x² – 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8,…}, D = {6}.
Solution.
Here A = {x : x ∈ R and x satisfies x² – 8x + 12 = 0}
= {x : x ∈ R and (x – 6)(x – 2) = 0} = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8,…….} and D = {6}
Now, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B and D ⊂ C

Miscellaneous Exercise Class 11 Maths Question 2.
In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B,then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 1

Miscellaneous Exercise Class 11 Maths Question 3.
Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B – C.
Solution.
Given that A ∩ B = A ∩C and A ∪ B=A ∪ C
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 2

Miscellaneous Exercise Class 11 Maths Question 4.
Show that the following four conditions are equivalent :
(i) A ⊂ B
(ii) A – B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution.
(i) ⇒ (ii)
A – B = {x: x∈ A and x∉B]
Since A⊂B
∴ A – B = φ

(ii) ⇒ (iii)
A – B = φ ⇒ A⊂B ⇒ A∪B = B

(iii) ⇒ (iv)
AuB = B ⇒A⊂B ⇒ A∩B = A

(iv) ⇒ (i)
A∩B = A ⇒ A⊂B
Thus (i) ⇔ (ii) ⇔ (iii) ⇔ (iv).

Miscellaneous Exercise Class 11 Maths Question 5.
Show that if A ⊂ B, then C – B ⊂ C – A.
Solution.
Let x ∈ C – B ⇒x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [∵ A ⊂ B]
⇒ x ∈ C – A
Hence C – B ⊂C – A

Miscellaneous Exercise Class 11 Maths Question 6.
Assume that P(A) = P(B). Show that A = B
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 3

Miscellaneous Exercise Class 11 Maths Question 7.
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B) Justify your answer.
Solution.
No, it is not true.
Take A = {1, 2} and B = {2,3}
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 4

Miscellaneous Exercise Class 11 Maths Question 8.
Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution.
(A ∩ B) ∪ (A – B) = (A ∩ B) ∪ (A ∩ B’)
= A ∩ (B ∪ B’) (By distributive law)
= A ∩ U = A
Hence, A = (A ∩ B) ∪ (A – B)
Also A ∪ (B – A) = A ∪ (B ∩ A’)
= (A ∪ B) ∩ (A ∪ A’) (By distributive law)
= (A ∪ B) ∩ U= A ∪ B
Hence, A ∪ (B – A) = A ∪ B.

Miscellaneous Exercise Class 11 Maths Question 9.
Using properties of sets, show that
(i) A ∪ (A ∩ B)=A
(ii) A ∩ (A ∪ B) = A.
Solution.
(i) We know that if A ⊂ B, then
A ∪ B = B. Also, A ∩ B ⊂ A
∴ A ∪ (A ∩ B) = A.
(ii) We know that if A ⊂ B,
then A ∩ B = A Also, A ⊂ A ∪ B
∴ A ∩ (A ∪ B) = A.

Miscellaneous Exercise Class 11 Maths Question 10.
Show that A ∩ B = A ∩ C need not imply B = C.
Solution.
Let A = {1, 2, 3, 4}, B = {2, 3, 4, 5, 6}, C = {2, 3, 4, 9,10}.
∴ A ∩ B = [1, 2,3,4} ∩ {2,3,4, 5, 6]
= {2, 3, 4}
A ∩ C = {1, 2, 3, 4} ∩ {2, 3, 4, 9, 10} = {2, 3, 4}
Now we have A ∩ B = A ∩ C. But B ≠ C.

Miscellaneous Exercise Class 11 Maths Question 11.
Let A and B be sets. If A ∩ X=B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ (A∪X), B = B ∩ (B ∪ X) and use Distributive law)
Solution.
Here
A ∪ X = B ∪ X for some set X
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 5

Miscellaneous Exercise Class 11 Maths Question 12.
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ
Solution.
Take A = {1, 2}, B = {1, 4} and C = {2, 4}
Now, A ∩ B = {1} ≠ φ, B ∩ C = {4} ≠ φ and
A ∩ C = {2} ≠ φ
But A ∩ B ∩ C = φ.

Miscellaneous Exercise Class 11 Maths Question 13.
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Solution.
Let T be the set of students who takes tea and C be the set of students who takes coffee. Here, n(T) = 150, n(C) = 225 and n(C ∩ T) = 100
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
= 150 + 225 -100 = 275
∴ Number of students taking either tea or coffee = 275
∴ Number of students taking neither tea nor coffee = 600 – 275 = 325.

Miscellaneous Exercise Class 11 Maths Question 14.
In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Solution.
Let H be the set of students who know Hindi and E be the set of students who know English.
Here n(H) = 100, n(E) = 50 and n(H ∩ E) = 25
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25 = 125

Miscellaneous Exercise Class 11 Maths Question 15.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper H, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T
and I, 3 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 6

Miscellaneous Exercise Class 11 Maths Question 16.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution.
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 8

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Chapter 1 Knowing Our Numbers Class 6 ncert solution

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers

Page No: 12

Exercise 1.1

1. Fill in the blanks:
(a) 1 lakh = _______ ten thousand.
(b) 1 million = _______ hundred thousand.
(c) 1 crore = _______ ten lakh.
(d) 1 crore = _______ million.
(e) 1 million = _______ lakh

Answer

a) 100
b) 10
c) 10
d) 10
e) 100

2. Place commas correctly and write the numerals:

(a) Seventy three lakh seventy five thousand three hundred seven.
(b) Nine crore five lakh forty one.
(c) Seven crore fifty two lakh twenty one thousand three hundred two.
(d) Fifty eight million four hundred twenty three thousand two hundred two.
(e) Twenty three lakh thirty thousand ten.

Answer

a) 73,75,307
b) 9,05,000,41
c) 7,52,21,302
d) 58,423,202
e) 23,30,010

3. Insert commas suitably and write the names according to Indian System of Numeration:

(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701

Answer

(a) 8,75,95,762: Eight crore seventy five lakh ninety five seven hundred sixty two.
(b) 85,46,283: Eighty five lakh forty six thousand two hundred eighty three
(c) 9,99,00,046: Nine crore ninety nine lakh forty six
(d) 9,84,32,701: Nine crore eighty four lakh thirty two thousand seven hundred one.

4. Insert commas suitably and write the names according to International System of Numeration:

(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831

Answer

(a) 78,921,092: Seventy eight million nine hundred twenty one thousand ninety two.
(b) 7,452,283: Seven million four hundred fifty two thousand two hundred eighty three.
(c) 99,985,102: Ninety nine million nine hundred eighty five thousand one hundred two.
(e) 48,049,831: Forty eight million forty nine thousand eight hundred thirty one.

Page No: 16

Exercise 1.2

1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Answer

Number of tickets sold first day = 1094
Number of tickets sold second day = 1812
Number of tickets sold third day = 2050
Number of tickets sold fourth day = 2751
Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707

2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Answer

Shekhar scored = 6980 runs
Shekhar wants to score = 10,000 runs
He need to score 10,000 – 6980 = 3020

3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?

Answer

Successful candidate registered 5,77,500 votes
Score secured by his rival = 3,48,700 votes

5,77,500 – 3,48,700 = 22880
Successful candidate need 22880 margin to win the election.

4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?

Answer

Books sold in the first week = 2,85,891
Books sold in the second week = 4,00,768
The sale of two weeks together = 2,85,891 + 4,00,768 = 686659
Second week of the month books sale 4,00,768 was greater than first week 2,85,891.

4,00,768 – 2,85,891 = 114877
Book sale was greater than by 114877

Page No: 17

5. Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once.

Answer

Greatest number = 76432
Smallest number = 23467
The difference between greatest and smallest number = 76432 – 23467 = 52965

6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Answer

Number of screw produced in one days = 2,825
As we know in the month of January there is 31 days.
Number of screw produced in 31 days = 2,825 × 31 = 87575.
So, the number of screw produced in January 2006 = 87575.

7. A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?

Answer

Merchant had 78,592 with her
Cost of one radio = 1200
Cost of 40 radio = 12000 × 40 = 48000
Money spent by merchant by = 48000
Money left = 78,592 – 48000 = 30592

Rs. 30592 will left after purchasing.

8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
(Hint: Do you need to do both the multiplications?)

Answer

Differences between 65 × 56 = 9
Correct answer is greater than by 7236 × 9 = 65124

9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
(Hint: convert data in cm.)

Answer

Cloth available = 40 m
As we know 1 m = 100 cm
40 m = 40 × 100 = 4000
2 m 15 cm = 215 cm
Required cloth for one t-shirt = 215 cm
Number of cloths stitched = 4000 ÷ 215

Therefore, 18 shirts can be made.
Remaining cloths is 1 m 130 cm.

10. Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Answer
Weight of each boxes is = 4 kg 500 g 1 kg = 1000 g4 kg 500 g = 4500 gSo, 800 kg = 800 × 1000 = 800000 g Number of boxes required for 800000 ÷ 4500

So, the number of boxes required for medicine = 177.
11. The distance between the school and the house of a student’s house is 1 km 875 m. Every day she walks both ways. Find the total distance covered by her in six days.
Answer
Distance between the school and the house = 1 km 875 m.As we know 1 km = 1000 m 1 km 875 m = 1875 m.There are two ways Distance covered by her in each day was 1875 × 2 = 3750 mAnd the distance covered in six days = 3750 × 6 = 22500∴ Distance covered in six days = 22500 m = 22 km 500 m .
12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Answer
Vessel can store = 4 litres and 500 mlAs we know, 1 l = 1000 ml We have to find number of classes that can store 25 ml of curd.Number of glasses = 4500 ÷25

∴ 180 glasses are required for 25 ml of curd.
Page No: 23
Exercise 1.3
1. Estimate each of the following using general rule: (a) 730 + 998 (b) 796 – 314 (c) 12,904 +2,888 (d) 28,292 – 21,496 Make ten more such examples of addition, subtraction and estimation of their outcome.
Answera) 730 + 998 • Rounding off to nearest hundred. 730 round off to 700 and 998 round off to 1000. 700 + 1000 = 1700
b) 796 – 314 800 – 300 = 500
c) 12,904 + 2,888 13000 + 3000 = 16000
d) 28,292 – 21,496 28000 – 21000 = 7000
2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) : (a) 439 + 334 + 4,317 (b) 1,08,734 – 47,599 (c) 8325 – 491 (d) 4,89,348 – 48,365 Make four more such examples.
Answer
a) 439 + 334 + 4,317 • Rounding of hundred = 400, 300, 4000400 + 300 + 4000 = 5000
• Rounding of ten = 440, 330, 4,320440 +330 + 4,320 = 5090.
b) 1,08,734 – 47,599 • Rounding of hundred = 1,08,700 and 47,6001,08,700 – 47,600 = 61100
• Rounding of ten = 1,08,730 and 47,6001,08,730 – 47,600
c) 8325 – 491 • Rounding of hundred 8300 and 5008300 – 500 = 7800
• Rounding of ten 8330 and 490 83330 – 490 = 7840
d) 4,89,348 – 48,365 • Rounding of hundred = 4,89,300 – 48400 = 440900• Rounding of ten = 489350 – 48370 = 440980
3. Estimate the following products using general rule:(a) 578 × 161 (b) 5281 × 3491 (c) 1291 × 592(d) 9250 × 29
Answer
a) Rounding off to nearest hundred ,578 round off to 600 and 161 round off to 200.600 × 200 = 120000
b) 5000 × 3000 = 15000000c) 1000 × 600 = 600000d) 9000 × 30 = 270000

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Class 11th chapter-1 unit & measurement | NCERT Solution | physics | class 11th physics solution

chapter-1 unit & measurement | NCERT Solution

Page No: 35

Exercises

2.1. Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to…..m³
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to … (mm)²
(c) A vehicle moving with a speed of 18 km h^–1covers….m in 1 s
(d) The relative density of lead is 11.3. Its density is ….g cm^–3or . …kg m^–3.

Answer

(a) Length of edge = 1cm = 1/100 m
Volume of the cube = side³
Putting the value of side, we get
Volume of the cube = (1/100 m)³

The volume of a cube of side 1 cm is equal to 10^-6 m³
(b) Given,
Radius, r = 2.0 cm = 20 mm (convert cm to mm)
Height, h = 10.0 cm =100 mm
The formula of total surface area of a cylinder S = 2πr (r + h)
Putting the values in this formula, we get
Surface area of a cylinder S = 2πr (r + h = 2 x 3.14 x 20 (20+100)
= 15072 = 1.5 × 10⁴ mm²
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.5 × 10⁴ mm²

(c) Using the conversion,
Given,
Time, t = 1 sec
speed = 18 km h^-1 = 18 km / hour
1 km = 1000 m and 1hour = 3600 sec
Speed = 18 × 1000 /3600 sec = 5 m /sec
Use formula
Speed = distance / time
Cross multiply it, we get
Distance = Speed × Time = 5 × 1 = 5 m
A vehicle moving with a speed of 18 km h^–1covers 5 m in 1 s.

(d) Density of lead = Relative density of lead × Density of water
Density of water = 1 g/cm³
Putting the values, we get
Density of lead = 11.3 × 1 g/ cm³
= 11.3 g cm^-3
1 cm = (1/100 m) =10^–2 m3
1 g = 1/1000 kg = 10^-3 kg
Density of lead = 11.3 g cm^-3 = 11.3
Putting the value of 1 cm and 1 gram
11.3 g/cm³ = 11.3 × 10^-3 kg (10^-2m)^-3 = 11.3 ×10^–3 × 10⁶ kg m^-3 =1.13 × 10³ kg m^–3
The relative density of lead is 11.3. Its density is 11.3 g cm^-3g cm^–3 or 1.13 × 10³ kg m^–3.

2.2. Fill in the blanks by suitable conversion of units:
(a) 1 kg m²s^–2= ….g cm² s^–2
(b) 1 m =….. ly
(c) 3.0 m s^–2=…. km h^–2
(d) G= 6.67 × 10^–11 N m² (kg)^–2=…. (cm)³s^–2 g^–1.

Answer

(a) 1 kg = 10³ g
1 m² = 10⁴ cm²
1 kg m² s^–2 = 1 kg × 1 m² × 1 s^–2
=10³ g × 10⁴ cm²× 1 s^–2 = 10⁷ g cm² s^–2
1 kg m²s^–2= 10⁷ g cm² s^–2

(b) Distance = Speed × Time
Speed of light = 3 × 10⁸ m/s
Time = 1 year = 365 days = 365 × 24 hours = 365 × 24 × 60 × 60 sec
Putting these values in above formula we get
1 light year distance = (3 × 10⁸ m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 10¹⁵ m
9.46 × 10¹⁵ m = 1 ly
So that 1 m = 1/ 9.46 × 10¹⁵ ly = 1.06 × 10^-16 ly

(c) 1 hour = 3600 sec so that 1 sec = 1/3600 hour
1 km = 1000 m so that 1 m = 1/1000 km
3.0 m s^–2 = 3.0 (1/1000 km)( 1/3600 hour)^-2 = 3.0 × 10^–3 km × ((1/3600)^-2 h–2)
= 3.0 × 10^–3 km × (3600)² h^–2 = 3.88 × 10⁴ km h^–2
3.0 m s^–2= 3.88 × 10⁴ km h^–2
(d) Given,
G= 6.67 × 10^–11 N m² (kg)^–2
We know that
1 N = 1 kg m s^–2
1 kg = 10³ g
1 m = 100 cm = 10² cm
Putting above values, we get
6.67 × 10^–11 N m² kg^–2 = 6.67 × 10^–11 × (1 kg m s^–2) (1 m²) (1Kg^–2)
Solve and cancel out the units we get
⇒ 6.67 × 10^–11 × (1 kg^–1 × 1 m³ × 1 s^–2)
Putting above values to convert Kg to g and m to cm
⇒ 6.67 × 10^–11 × (10³ g)^-1 × (10² cm)³ × (1 s^–2)
⇒ 6.67 × 10^–11 × 10^-3 g^-1 × 10⁶ cm³ × (1 s^–2)
⇒ 6.67 × 10^–8 cm³ s^–2 g^–1
G= 6.67 × 10^–11 N m² (kg)^–2= 6.67 × 10^–8 (cm)³s^–2 g^–1.
2.3. A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m²s^–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α^–1 β^–2 γ²in terms of the new units.

Answer

Given that,
1 Calorie=4.2 J = 4.2 Kg m² s^-2 …… (i)
As new unit of mass = α Kg
∴ 1 Kg = 1/α new unit of mass
Similarly, 1 m = β^-1 new unit of length
1 s = γ^-1 new unit of time
Putting these values in (i), we get
1 calorie = 4.2 (α^-1 new unit of mass) (β^-1 new unit of length)2 (γ^-1 new unit of time)-2
= 4.2 α^-1 β^-2 γ² new unit of energy (Proved)

2.4. Explain this statement clearly:“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light.
Answer

The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.
(a) An atom is a very small object in comparison to a soccer ball.
(b) A jet plane moves with a speed greater than that of a bicycle.
(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.
(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.
(e) A proton is more massive than an electron.
(f) Speed of sound is less than the speed of light.

2.5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer

Distance between the Sun and the Earth = Speed of light x Time taken by light to cover the distance
Given that in the new unit, speed of light = 1 unit
Time taken, t = 8 min 20 s = 500 s
∴Distance between the Sun and the Earth = 1 x 500 = 500 units

2.6. Which of the following is the most precise device for measuring length:
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?

Answer
(a) Least count of this vernier callipers = 1SD – 1 VD = 1 SD – 19/20 SD = 1/20 SD
= 1.20 mm = 1/200 cm = 0.005 cm

(b) Least count of screw gauge = Pitch/Number of divisions = 1/1000 = 0.001 cm.

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

2.7. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?
Answer
Magnification of the microscope = 100
Average width of the hair in the field of view of the microscope = 3.5 mm
∴Actual thickness of the hair is 3.5/100 = 0.035 mm.
2.8. Answer the following:

(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?
Answer
Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,
Diameter = Length of thread/Number of turns

(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
Answer
It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.

(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer

A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

2.9. The photograph of a house occupies an area of 1.75 cm²on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement?

Answer

Area of the house on the slide = 1.75 cm²
Area of the image of the house formed on the screen = 1.55 m²= 1.55 × 104 cm²
2.10. State the number of significant figures in the following:

(a) 0.007 m²► 1

(b) 2.64 x 10²⁴ kg
► 3

(d) 6.320 J
► 4

(e) 6.032 N m^-2
► 4

(f) 0.0006032 m²
► 4
Page No: 36

2.11. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Answer

Given that,
length, l = 4.234 m
breadth,b = 1.005 m
thickness, t = 2.01 cm = 2.01 × 10^-2 m
Area of the sheet = 2 (l × 0 + b × t + t × l) = 2 (4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 2 (4.3604739) = 8.7209478 m2
As area can contain a maximum of three significant digits, therefore, rounding off, we get
Area = 8.72 m²
Also, volume = l × b × t
V = 4.234 × 1.005 × 0.0201 = 0.0855289 = 0.0855 m³ (Significant Figures = 3)

2.12. The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Answer

Mass of grocer’s box = 2.300 kg
Mass of gold piece I = 20.15g = 0.02015 kg
Mass of gold piece II = 20.17 g = 0.02017 kg

(a) Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg
In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg.

(b) Difference in masses = 20.17 – 20.15 = 0.02 g
In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

2.13. A physical quantity P is related to four observables a, b, c and d as follows:
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Answer

Percentage error in P = 13 %
Value of P is given as 3.763.
By rounding off the given value to the first decimal place, we get P = 3.8.

2.14. A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Answer

The displacement y has the dimension of length, therefore, the formula for it should also have the dimension of length. Trigonometric functions are dimensionless and their arguments are also dimensionless. Based on these considerations now check each formula dimensionally.
The formulas in (ii) and (iii) are dimensionally wrong.

2.15. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :
m = m₀/ (1-v²)^1/2
Guess where to put the missing c.

Answer

Given the relation,
m = m₀/ (1-v²)^1/2
Dimension of m = M¹ L⁰ T⁰
Dimension of m₀ = M¹ L⁰ T⁰
Dimension of v = M⁰ L¹ T^–1
Dimension of v² = M⁰ L² T^–2
Dimension of c = M⁰ L¹ T^–1
The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, (1-v²)^1/2 is dimensionless i.e., (1 – v²) is dimensionless. This is only possible if v² is divided by c². Hence, the correct relation is
m = m₀/ (1 – v²/c²)^1/2
2.16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10^-10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms ?

Answer

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10^-10m
Volume of hydrogen atom = (4/3) π r³
= (4/3) × (22/7) × (0.5 × 10^-10)³
= 0.524 × 10^-30 m³
1 mole of hydrogen contains 6.023 × 10²³ hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms = 6.023 × 10²³ × 0.524 × 10^–30
= 3.16 × 10^–7 m³
2.17. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ?

Answer

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10^-10 m
Volume of hydrogen atom = (4/3) π r³
= (4/3) × (22/7) × (0.5 × 10^-10)³
= 0.524 × 10^-30 m³
Now, 1 mole of hydrogen contains 6.023 × 10²³ hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms, V_a = 6.023 × 10²³ × 0.524 × 10^–30
= 3.16 × 10^–7 m³
Molar volume of 1 mole of hydrogen atoms at STP,
V_m = 22.4 L = 22.4 × 10^–3 m³

Hence, the molar volume is 7.08 × 10⁴ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

2.18. Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer

Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.
On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

2.19. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 10¹¹m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1^“ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1^“ (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of meters?

Answer

Diameter of Earth’s orbit = 3 × 10¹¹ m
Radius of Earth’s orbit, r = 1.5 × 10¹¹ m
Let the distance parallax angle be 1″ = 4.847 × 10^–6 rad.
Let the distance of the star be D.
Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1″

= 0.309 × 10^-6 ≈ 3.09 × 10¹⁶ m
Hence, 1 parsec ≈ 3.09 × 10¹⁶ m

2.20. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Answer

Distance of the star from the solar system = 4.29 ly
1 light year is the distance travelled by light in one year.
1 light year = Speed of light × 1 year
= 3 × 10⁸ × 365 × 24 × 60 × 60 = 94608 × 10¹¹ mhttps://6888dfb24f93374e0a2a625caf1d8094.safeframe.googlesyndication.com/safeframe/1-0-38/html/container.html∴ 4.29 ly = 405868.32 × 10¹¹ m
∵ 1 parsec = 3.08 × 10¹⁶ m
∴ 4.29 ly = 405868.32 × 10¹¹ / 3.08 × 10¹⁶ = 1.32 parsec
Using the relation,
θ = d / D
where,
Diameter of Earth’s orbit, d = 3 × 10¹¹ m
Distance of the star from the earth, D = 405868.32 × 10¹¹ m
∴ θ = 3 × 10¹¹/ 405868.32 × 10¹¹ = 7.39 × 10^-6 rad
But, 1 sec = 4.85 × 10^–6 rad
∴ 7.39 × 10^-6 rad = 7.39 × 10^-6/ 4.85 × 10^-6= 1.52^“
2.21. Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer

It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ∼ 10^–15 s) are used to measure time intervals in several physical and chemical processes.
X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.

The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

2.22. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.

Answer

(a) During monsoons, a Metrologist records about 215 cm of rainfall in India i.e., the height of water column, h = 215 cm = 2.15 m
Area of country, A = 3.3 × 10¹² m²
Hence, volume of rain water, V = A × h = 7.09 × 10¹² m³
Density of water, ρ = 1 × 10³ kg m^–3
Hence, mass of rain water = ρ × V = 7.09 × 10¹⁵ kg
Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 10¹⁵ kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say d₁).
Volume of water displaced by the ship, V_b = Ad₁
Now, move an elephant on the ship and measure the depth of the ship (d₂) in this case.
Volume of water displaced by the ship with the elephant on board, V_be= Ad₂
Volume of water displaced by the elephant = Ad₂ – Ad₁
Density of water = D
Mass of elephant = AD (d₂ – d₁)

(c) Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.

(d) Area of the head surface carrying hair = A
With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.
∴Area of one hair = πr²
Number of strands of hair ≈ Total surface area / Area of one hair = A / πr²
(e) Let the volume of the room be V.
One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10^–3 m³ volume.
Number of molecules in one mole = 6.023 × 10²³
∴Number of molecules in room of volume V
= 6.023 × 10²³ × V / 22.4 × 10^-3 = 134.915 × 10²⁶ V = 1.35 × 10²⁸ V

2.23. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 10³⁰ kg, radius of the Sun = 7.0 × 10⁸ m.

Answer

Mass of the Sun, M = 2.0 × 10³⁰ kg
Radius of the Sun, R = 7.0 × 10⁸ m
Density, ρ = ?

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.

2.24. When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72^“ of arc. Calculate the diameter of Jupiter.

Answer

Distance of Jupiter from the Earth, D = 824.7 × 10⁶ km = 824.7 × 10⁹ m
Angular diameter = 35.72^“ = 35.72 × 4.874 × 10^-6 rad
Diameter of Jupiter = d
Using the relation,
θ = d/ D
d = θ D = 824.7 × 10⁹ × 35.72 × 4.872 × 10^-6
= 143520.76 × 10³ m = 1.435 × 10⁵ Km

Additional Exercises

2.25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v : tan θ = v and checks that the relation has a correct limit: as v → 0, θ → 0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.

Answer

Incorrect; on dimensional ground
The relation is tan θ = ν
Dimension of R.H.S = M⁰ L¹ T^–1
Dimension of L.H.S = M⁰ L⁰ T⁰
(∵ The trigonometric function is considered to be a dimensionless quantity)
Dimension of R.H.S is not equal to the dimension of L.H.S. Hence, the given relation is not correct dimensionally.
To make the given relation correct, the R.H.S should also be dimensionless. One way to achieve this is by dividing the R.H.S by the speed of rainfall ν’
Therefore, the relation reduces to
tan θ = ν / ν’
This relation is dimensionally correct.

2.26. It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?

Answer

Error in 100 years = 0.02 s

Hence, the accuracy of a standard caesium clock in measuring a time interval of 1 s is 10^-12 s.

2.27. Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase : 970 kg m^-3. Are the two densities of the same order of magnitude ? If so, why ?

Answer

Diameter of sodium atom = Size of sodium atom = 2.5 Å
Radius of sodium atom, r = (1/2) × 2.5 Å = 1.25 Å = 1.25 × 10^-10 m
Volume of sodium atom, V = (4/3) π r³
= (4/3) × 3.14 × (1.25 × 10^-10)³ = V_Sodium
According to the Avogadro hypothesis, one mole of sodium contains 6.023 × 10²³ atoms and has a mass of 23 g or 23 × 10^–3 kg.
∴ Mass of one atom = 23 × 10^-3/ 6.023 × 10²³ Kg = m₁
Density of sodium atom, ρ = m₁ / V_Sodium
Substituting the value from above, we get
Density of sodium atom, ρ =4.67 × 10^-3Kg m^-3
It is given that the density of sodium in crystalline phase is 970 kg m^–3.

Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.

2.28. The unit of length convenient on the nuclear scale is a Fermi : 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation :
r = r₀ A^1/3
where r is the radius of the nucleus, A its mass number, and r₀ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Answer

Radius of nucleus r is given by the relation,
r = r₀ A^1/3
r₀ = 1.2 f = 1.2 × 10^-15 m
Volume of nucleus, V = (4 / 3) π r³
= (4 / 3) π (r₀A^1/3)³ = (4 / 3) π r₀A ….. (i)

Now, the mass of a nuclei M is equal to its mass number i.e.,
M = A amu = A × 1.66 × 10^–27 kg
Density of nucleus, ρ = Mass of nucleus / Volume of nucleus
= A X 1.66 × 10^-27/ (4/3) π r₀³ A

= 3 X 1.66 × 10^-27/ 4 π r₀³ Kg m^-3
his relation shows that nuclear mass depends only on constant r₀. Hence, the nuclear mass densities of all nuclei are nearly the same.
Density of sodium nucleus is given by,
ρ_Sodium = 3 × 1.66 × 10^-27/ 4 × 3.14 × (1.2 × 10^-15)³
= 4.98 × 10¹⁸/ 21.71 = 2.29 × 10¹⁷ Kg m^-3
2.29. A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?

Answer

Time taken by the laser beam to return to Earth after reflection from the Moon = 2.56 s
Speed of light = 3 × 10⁸ m/s
Time taken by the laser beam to reach Moon = 1 / 2 × 2.56 = 1.28 s
Radius of the lunar orbit = Distance between the Earth and the Moon = 1.28 × 3 × 10⁸ = 3.84 × 10⁸ m = 3.84 × 10⁵ km

2.30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine?
(Speed of sound in water = 1450 m s^-1).

Answer

Let the distance between the ship and the enemy submarine be ‘S’.
Speed of sound in water = 1450 m/s
Time lag between transmission and reception of Sonar waves = 77 s
In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine (2S).
Time taken for the sound to reach the submarine = 1/2 × 77 = 38.5 s
∴ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km

2.31. The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?

Answer

Time taken by quasar light to reach Earth = 3 billion years
= 3 × 10⁹ years
= 3 × 10⁹ × 365 × 24 × 60 × 60 s
Speed of light = 3 × 10⁸ m/s
Distance between the Earth and quasar
= (3 × 10⁸) × (3 × 10⁹ × 365 × 24 × 60 × 60)
= 283824 × 10²⁰ m
= 2.8 × 10²² km

2.32. It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Answer

The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.
Position of sun moon during lunar eclipse
Distance of the Moon from the Earth = 3.84 × 10⁸ m
Distance of the Sun from the Earth = 1.496 × 10¹¹ m
Diameter of the Sun = 1.39 × 10⁹ m

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:
PQ / RS = VT / UT
1.39 x 10⁹/ RS = 1.496 × 10¹¹/ 3.84 × 10⁸
RS = (1.39 × 3.84 / 1.496) × 10⁶ = 3.57 × 10⁶ m
Hence, the diameter of the Moon is 3.57 × 10⁶ m.

2.33. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

Answer

One relation consists of some fundamental constants that give the age of the Universe by:

t = (e²/4πε₀)² × (1 / m_pm_e²c³G)

Where,
t = Age of Universe
e = Charge of electrons = 1.6 ×10^–19 C
ε₀ = Absolute permittivity
m_p= Mass of protons = 1.67 × 10^–27 kg
m_e= Mass of electrons = 9.1 × 10^–31 kg
c = Speed of light = 3 × 10⁸ m/s
G = Universal gravitational constant = 6.67 × 10¹¹ Nm² kg^–2
Also, 1 / 4πε₀ = 9 × 10⁹Nm²/C²
Substituting these values in the equation, we get
t = (1.6 × 10^-19)⁴ × (9 × 10⁹)² / (9.1 × 10^-31)² × 1.67 × 10^-27 × (3 × 10⁸)³ × 6.67 × 10^-11
= [ (1.6)⁴× 81 / 9.1 × 1.67 × 27 × 6.67 ] × 10^{-76+18-62+27-24+11} seconds= [(1.6)⁴ × 81 / 9.1 × 1.67 × 27 × 6.67 × 365 × 24 × 3600 ] × 10^{-76+18+62+27-24+11} years≈ 6 X 10^-9 × 10¹⁸ years= 6 billion years.

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Ch 8 Challenges to Democracy Class 10 Important Questions | NCERT Social- Science Chapter-8 – Edu grown

Question 1.
Describe in brief the three challenges faced by democracy. (2014 D)
Answer:

Foundational challenge. It relates to making the transition to democracy and then instituting democratic government. It involves bringing down the existing non-democratic regime, keeping military away from controlling government and establishing a sovereign and functional State.
Challenge of expansion. It involves applying the basic principle of democratic government across all the regions, different social groups and various institutions. It pertains to ensuring greater power to local governments, extension of federal principle to all the units of the federation, inclusion of women and minority groups, etc. Most established democracies, e.g., India and US, face the challenge of expansion.
Challenge of deepening of democracy. This challenge involves strengthening of the institutions and practices of democracy. It means strengthening those institutions that help people’s participation and control in the government. It aims at bringing down the control and influence of rich and powerful people in making governmental decisions.

Question 2.
What do you mean by foundational challenge in democracy? What values can help to overcome this challenge? (2014 OD)
Answer:
Transition to democratic institutions from non-democratic regimes, separation of military from governing authority, establishing a sovereign and a functional state can be some of the foundational challenges in democracies.
The values that may help overcome them are:

honesty
equality
freedom

Question 3.
Describe in brief the three challenges faced by democracy.
Answer:

Foundational challenge. It relates to making the transition to democracy and then instituting democratic government. It involves bringing down the existing non-democratic regime, keeping military away from controlling government and establishing a sovereign and functional State.
Challenge of expansion. It involves applying the basic principle of democratic government across all the regions, different social groups and various institutions. It pertains to ensuring greater power to local governments, extension of federal principle to all the units of the federation, inclusion of women and minority groups, etc. Most established democracies, e.g., India and US, face the challenge of expansion.
Challenge of deepening of democracy. This challenge involves strengthening of the institutions and practices of democracy. It means strengthening those institutions that help people’s participation and control in the government. It aims at bringing down the control and influence of rich and powerful people in making governmental decisions.

Question 4.
Explain with examples why some laws that seek to ban something are not very successful in politics. (2011 D)
Answer:
Law has an important role to play in political reform. Carefully devised changes in law can help to discourage wrong political practices and encourage good ones. But legal constitutional changes by themselves are not effective, until carried out by political activists, parties, movements and politically conscious citizens. Any legal change must carefully look at what results it will have on politics. Sometimes it can be counter-productive.

For example, many states have banned people who have more than two children from contesting panchayat elections. This has resulted in denial of democratic opportunity to many poor men and women.

The best laws are those which empower the people to carry out democratic reforms. The Right to Information Act is a good example that supplements the existing laws. “Any law for political reforms is a good solution but who will implement it and how”—is the question. It is not necessary that the legislators will pass legislations that go against the interests of the political parties and MPs.

Question 5.
“Legal constitutional changes by themselves cannot overcome challenges to democracy.” Explain with example. (2015 D, 2013 D, 2011 D)
Or
How are the challenges to democracy linked to the possibility of political reforms? Explain.
Or
Suggest any five political reforms to strengthen democracy. (2014 D)
Answer:
As legal constitutional changes by themselves cannot overcome challenges to democracy, democratic reforms need to be carried out mainly by political activists, parties, movements and politically conscious citizens.
(i) Any legal change must carefully look at what results it will have on politics. Generally, laws that seek a ban on something are rather counter-productive;
For example, many states have debarred people who have more than two children from contesting Panchayat elections. This has resulted in denial of democratic opportunity to many poor women, which was not intended. The best laws are those which empower people to carry out democratic reforms; for example, the Right to Information Act which acts as a watchdog of democracy by controlling corruption.

(ii) Democratic reforms are to be brought about principally through political parties. The most important concern should be to increase and improve the quality of political participation by ordinary citizens.

(iii) Any proposal for political reforms should think not only about what is a good solution, but also about who will implement it and how. Measures that rely on democratic movements, citizens’ organizations and media are likely to succeed.

Question 6.
Explain the ‘foundational challenge’ of democracy by stating three points. (2011 D)
Answer:

Foundational challenge relates to making the transition to democracy and then instituting democratic government. It involves establishing a sovereign and functional state.
It involves bringing down the existing non-democratic regime, keeping military away from controlling government and establishing a civilian control over all governmental institutions by holding elections.
It involves the recognition of people’s choice and opportunity to change rulers, recognise people’s will. In countries like Myanmar political leader Suu Kyi has been kept under house arrest for more than 20 years. Thus, in this case, foundational challenge recognizes the need to release political leaders and recall them from exile and holding of multiparty elections.

Question 7.
Explain ‘the challenge of deepening of democracy’ by stating three points. (2012 D, 2014 OD)
Answer:
The challenge of deepening of democracy:

This challenge involves strengthening of the institutions and practices of democracy. It means strengthening those institutions that help people’s participation and control in the government.
The challenge lies in realising the expectations of the people in a democracy. It is possible that some significant decisions may take place through consensus but challenging moments in democracy usually involve conflict between those groups who have power and those who aspire for a share in power. In Bolivia, the water struggle was a challenge of deepening of democracy.
The challenge of deepening of democracy is faced by every nation in one form or another. It aims at bringing down the control and influence of the rich and powerful people in making governmental decisions. The need is for individual freedom and dignity to have legal and moral force.

Question 8.
How are some countries of the world facing the ‘challenge of expansion of democracy’? Explain with examples. (2012 D, 2012 OD)
“Most of the established democracies are facing the challenge of expansion.” Support the statement with examples. (2016 D)
Answer:
Most of the established democracies face the challenge of expansion. This involves applying the basic principle of democratic government across all the regions, different social groups and various institutions. Ensuring greater power to local government, extension of federal principle to all the units of federation, inclusion of women and minority groups, etc. falls under this challenge. This means less and less decisions should remain outside the arena of democratic control. Most of the countries including India and the US face this challenge.

Question 9.
Explain with examples how do some countries face foundational challenge of democracy. (2013 OD)
Answer:

Foundational challenge relates to making the transition to democracy and then instituting democratic government. It involves establishing a sovereign and functional state.
It involves bringing down the existing non-democratic regime, keeping military away from controlling government and establishing a civilian control over all governmental institutions by holding elections. Eg: Nepal, Egypt, Pakistan.
In countries like Pakistan, democracy comes for and/ or remains for a short time and gets replaced by dictatorial rule.
It involves the recognition of people’s choice and opportunity to change rulers, recognize people’s will. In countries like Myanmar political leader Suu Kyi was kept under house arrest for more than 20 years. Thus, in this case, foundational challenge recognizes the need to release political leaders and recall them from exile and holding of multiparty elections.

Question 10.
Analyse three major challenges before countries which do not have democratic form of governments. (2013 OD)
Answer:
Challenges faced by countries which do not have a democratic form of government:

These countries face the foundational challenge of making the transition to democracy and then instituting democratic government.
They also face the challenge of bringing down the existing non-democratic regime, and keeping the military away from controlling the government.
Such countries have to make great efforts to establish a sovereign and functional State

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Chapter 7 Outcomes of Democracy Class 10 Important Questions | NCERT Social- Science Chapter-7 – Edu grown

Question 1.
Analyse any three values that make democracy better. (2017 D)
Answer:
We feel that democracy is a better form of government than any other form of government because:

Democracy promotes equality among citizens.
It enhances dignity of individual. It promotes dignity of women and strengthens the claims of the disadvantaged.
It improves the quality of decision making. There is transparency in a democracy.
It provides methods to resolve conflicts.
Democracy allows room to correct mistakes.

Question 2.
On the basis of which values will it be a fair expectation that democracy should produce a harmonious social life? Explain. (2017 OD)
Answer:
No society can fully and permanently resolve conflicts among different groups. But we can certainly learn to respect these differences and evolve a mechanism to negotiate the differences. Belgium is an example of how successfully differences were negotiated among ethnic groups. Therefore, democracy is best suited to accommodate various social divisions as it usually develops a procedure to conduct their competition. But the example of Sri Lanka shows how distrust between two communities turned into widespread conflict. Thus, a democracy must fulfil the following conditions and be based on these values in order to achieve a harmonious social life—

Majority and minority opinions are not permanent. Democracy is not simply rule by majority opinion. The majority needs to work with minority so that government may function to represent the general view.
Rule by majority does not become rule by majority community in terms of religion or race or linguistic groups, etc.
Democracy remains democracy so long as every citizen has a chance of being in majority at some point of time. No individual should be debarred from participating in a democracy on the basis of religion, caste, community, creed and other such factors.

Question 3.
Why do we feel that democracy is a better form of government than any other form? Explain. 2015OD Answer: Democracy is a better form of government than any other form because:

It is based on the idea of deliberation and negotiation. Thus the necessary delay in implementation.
Decisions are acceptable to people and are more effective.
A citizen has the right and the means to examine the process of decision-making. There is transparency in a democracy.
Democratic government is a legitimate government, people’s own government.
Ability to handle differences, decisions and conflicts is a positive point of democratic regimes.
Democracy has strengthened the claims of the disadvantaged and discriminated castes for equal status and equal opportunity.

Question 4.
Why do we feel that democracy is a better form of government than any other form of government? Explain. (2012 OD)
Or
How do you feel that democracy is better than any other form of government? Explain. (2013 OD)
Or
“Democracy is more effective than its other alternatives.” Justify the statement. (2015 D)
Answer:
We feel that democracy is a better form of government than any other form of government because:

Democracy promotes equality among citizens.
It enhances dignity of individual. It promotes dignity of women and strengthens the claims of the disadvantaged.
It improves the quality of decision making. There is transparency in a democracy.
It provides methods to resolve conflicts.
Democracy allows room to correct mistakes.

Question 5.
How do democracies accommodate social diversity? Explain with examples. (2011 OD, 2014 OD)
Or
Explain the conditions in which democracies are able to accommodate social diversities. (2012 D)
Or
“Democracies lead to peaceful and harmonious life among citizens”. Support the statement with suitable examples. (2013 OD)
Answer:
No society can fully and permanently resolve conflicts among different groups. But we can certainly learn to respect these differences and evolve a mechanism to negotiate the differences. Belgium is an example of how successfully differences were negotiated among ethnic groups. Therefore, democracy is best suited to accommodate various social divisions as it usually develops a procedure to conduct their competition. But the example of Sri Lanka shows how distrust between two communities turned into widespread conflict, and thus a democracy must fulfil the following two conditions in order to achieve a harmonious social life:

Majority and minority opinions are not permanent. Democracy is not simply rule by majority opinion. The majority needs to work with minority so that government may function to represent the general view.
Rule by majority does not become rule by majority community in terms of religion or race or linguistic groups, etc.
Democracy remains democracy so long as every citizen has a chance of being in majority at some point of time. No individual should be debarred from participating in a democracy on the basis of religion, caste, community, creed and other such factors.

Question 6.
Explain the ways in which democracy has succeeded in maintaining dignity and freedom of citizens. (2012 D)
Or, “Democracy stands much superior to any other form of government in promoting dignity and freedom of the individual.” Support the statement with suitable examples. 20130D
Answer:
The passion of respect and freedom are the basis of democracy:

Economic disparity in society has been minimized to a great extent.
In many democracies women were deprived of their right to vote for a long period of time. After long struggle they achieved their right, respect and equal treatment.
Democracy in India has strengthened the claims of the disadvantaged and discriminated castes for equal states and opportunities, for example, SCs and STs.
In democracy all adult citizens have the right to vote.
Democracy evolves a mechanism that takes into account the differences and intrinsic attributes of various ethnic groups. In a democracy majority always needs to work taking into account the interest of the minority so that the minority do not feel alienated.

Question 7.
How is democracy a better form of government in comparison with other forms of governments? Explain. (2016 D, 2014 D)
Or, “There is an overwhelming support for the idea of democracy all over the world.” Support the statement. (2015 OD)
Answer:
Over a hundred countries of the world today claim and practice some kind of democratic politics.

They have formal constitutions, hold elections, have parties and they guarantee rights of citizens.
Thus, in most countries, the democracy produces a government that is accountable to the citizens and responsive to the needs and expectations of the citizens.
No society can fully and permanently resolve conflict among different groups. But we can learn to respect these differences and evolve mechanisms to negotiate them. Democracy is best suited as it develops a procedure to conduct competitions. Belgium is a successful example of negotiating difference among ethnic population.
Passion for respect and freedom is the basis of democracy and has been achieved in various degrees in various democracies.
The support for democracy is overwhelming all over the world and is evident from South Asia, where the support exists in countries with democratic as well as undemocratic regimes.
People wish to be ruled by representatives elected by them as a democratic government is people’s own government and makes them believe that it is suitable for their country as it is a legitimate government.

Question 8.
“Most destructive feature of democracy is that its examination never gets over.” Support the statement with appropriate arguments. (2011 D)
Answer:
Suitable arguments:

As people get some benefits of democracy, they ask for more.
People always come up with more expectations from the democratic set up.
They also have complaints against democracy.
More and more suggestions and complaints by the people is also a testimony to the success of democracy.
A public expression of dissatisfaction with democracy shows the success of the democratic project.

Question 9.
“Democracy is seen to be good in principle but felt to be not so good in practice.” Justify the statement. (2013 D)
Answer:
If we look at some of the democratic policies being implemented in more than one hundred countries of the world, democracy seems to be good. For example, having a formal Constitution, holding regular elections, guaranteeing the citizens certain rights, working for the welfare of the people etc. make us advocate that democracy is good.

But if we look in terms of social situations, their economic achievements and varied cultures, we find a very big difference in most of the democracies. The vast economic disparities, social injustice based on discrimination, standard of life, sex discrimination, etc. create many doubts about the merits of democracy. Whenever some of our expectations are not met, we start blaming the idea of democracy. Since democracy is a form of government, it can only create conditions for achieving our goals if they are reasonable.

Question 10.
“Democracy stands much superior in promoting dignity and freedom of the citizens”. Justify the statement. (2016 OD)
Answer:
Examples to illustrate that dignity and freedom of citizens are best guaranteed in a democracy:

(i) Dignity of women. Democracy recognizes dignity of women as a necessary ingredient of society. The one way to ensure that women related problems get adequate attention is to have more women as elected representatives. To achieve this, it is legally binding to have a fair proportion of women in the elected bodies. Panchayati Raj in India has reserved one-third seats in local government bodies for women. In March 2010, the Women’s Reservation Bill was passed in the Rajya Sabha ensuring 33% reservation for women in Parliament and State legislative bodies.

(ii) Democracy has strengthened the claims of disadvantaged and discriminated castes. When governments are formed, political parties usually take care that representatives of different castes and tribes find a place in it. Some political parties are known to favour some castes. Democracy provides for equal status and opportunities for all castes.

(iii) Democracy transforms people from the status of a subject into that of a citizen. A democracy is concerned with ensuring that people will have the right to choose their rulers and people will have control over the rulers. Whenever possible and necessary, citizens should be able to participate in decision-making that affects them all.

(iv) A citizen has the right and the means to examine the process of decision-making. There is transparency in a democracy like India. In October 2005, the Right to Information (RTI) law was passed which ensures all its citizens the right to get all the information about the functions of the government departments. In a democracy, people also have the right to complain about its functioning.

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Chapter 6 Political Parties Class 10 Important Questions | NCERT Social- Science Chapter-6- Edu grown

Question 1.
“No party system is ideal for all countries and in all situations.” Justify the statement with five arguments. (2013 0D)
Answer:
Parties are a necessary condition for a democracy. The rise of political parties is directly linked to the emergence of representative democracies. Party system is not something any country can choose. It evolves depending on the nature of society, its social and regional divisions, its history of politics and system of elections.

Each country develops a party system that is conditioned by its special circumstances.

For example, India has evolved a multi-party system, because of its social and geographical diversity which cannot be easily absorbed by two or three parties.
Political parties make policies to promote collective good and there can be different views on what is good for all. Therefore no system is ideal for all countries and situations.

Question 2.
Describe the three components of a political party. (2014 D, 2015 OD)
Answer:
Components of a political party are:

The leaders,
active members and
the followers.
- The leaders are recruited and trained by parties. They are made ministers to run the government. The big policy decisions are taken by the political executives that come from the political parties.
- Parties have lakhs of members and activists spread over the country. Many of the pressure groups are the extensions of political parties among different sections of society. But since most of the members belong to a party, they go by the direction of the party leadership, irrespective of their personal opinion.
- The followers are the ordinary citizens, who believe in the policies of their respective party and give popular support through elections. Often the opinion of the followers crystallise on the lines parties take.

Question 3.
What is a multi-party system? Why has India adopted a multi-party system? Explain. (2015 D)
Or
How has multi-party system strengthened democracy in India? (2012 D)
Answer:
Multi-party system. In this system, the government is formed by various parties coming together in a coalition. When several parties in a multi-party system join hands for the purpose of contesting elections and winning power, it is called an alliance or a front.

For example, in India there were three major alliances in 2004 parliamentary elections—the National Democratic Alliance (NDA), the United Progressive Alliance (UPA) and the Left Front. This system on one hand leads to political instability but at the same time, allows for a variety of interests and opinions to enjoy political representation.

Each country develops a party system that is suitable for its special circumstances. India has evolved as a multi-party system because its social and geographical diversity cannot be absorbed by two or three parties. Thus, such representation strengthens democracy. Multi-party system facilitates representation of regional and geographical diversities. In India, several regional parties are in power at the State level such as the DMK in Tamil Nadu, Akali Dal in Punjab the BSP in Uttar Pradesh.

Question 4.
“Lack of internal democracy within parties is the major challenge to political parties all over the world”. Analyse the statement. (2015 D)
Answer:

Most political parties do not practise open and transparent procedures for their functioning like maintaining membership registers, holding organisational meetings or conducting internal elections regularly. Thus, ordinary members of the party do not get sufficient information on the happenings in the party and have no means to influence the decisions.
Also, there are very few chances for an ordinary worker to rise to the top in a party. Since one or, at the most, a few leaders exercise paramount power in the party, those who disagree with the leadership, find it difficult to continue in the party.
Those who happen to be the leaders are in a position to take undue advantage and favour people close to them or even their family members. And, in many parties, the top positions are invariably controlled by members of one family which is bad for democracy.

Question 5.
“Dynastic succession is one of the most serious challenges before the political parties.” Analyse the statement. (2015 OD)
Answer:
Most political parties do not practise open and transparent procedures for their functioning. So there are very few ways for an ordinary worker to rise to the top in a party. Those who happen to be the leaders are in a position of unfair advantage as they favour people close to them or even their family members. In many parties in India, we see a trend of dynastic succession. The top positions are always controlled by members of a particular family, which is unfair to other members of the party, and bad for democracy. This is so because people who do not have adequate experience or popular support come to occupy positions of power.

More than loyalty to party principles and policies, personal loyalty to the leader becomes more important. This tendency is seen all over the world, even in older democracies.

Question 6.
What is meant by a ‘national political party’? State the conditions required to be a national political party. (2016 D)
Answer:
National political parties have their units in various states. By and large all these units follow the same policies, programmes and strategy that is decided at the national level.
Conditions required to be a national political party:

A party that secures at least 6% of the total votes in general elections of Lok Sabha or assembly elections in four states.
A party that wins at least 4 seats in the Lok Sabha.

Question 7.
What is meant by regional political party? State the conditions required to be recognised as a ‘regional political party’. (2016 OD)
Answer:
A regional party is a party that is present in only some states. Regional parties or State parties need not be regional in their ideology. They have state identity as they are present only in some states. Some of these parties are all India parties that happen to have succeeded only in states. Example, Samajwadi Party, Rashtriya Janta Dal.
Conditions required for a party to be recognized as a regional political party:

A party that secures atleast six percent of the total votes in an election to the legislative assembly of a state.
Wins atleast two seats in the legislative assembly.

Question 8.
“Nearly every one of the state parties wants to get an opportunity to be a part of one or the other national level coalition.” Support the statement with arguments. (2016 D)
Answer:
The state parties also referred to as regional parties are not necessarily regional in ideology. Some of these parties are all India parties that happen to have succeeded only in some states. Over the last three decades, the number and strength of these parties has expanded. Before the general elections one national party was able to secure on its own a majority in the Lok Sabha.

As a result, the national parties were compelled to form alliances with state parties.

Since 1996, nearly every one of the state parties got an opportunity to be a part of one or the other national level coalition government. This contributed to the strengthening of federalism and democracy. Example of state parties having national level political organisation with units in several states are Samajwadi Party (SP), Rashtriya Janata Dal, Samata Party.

Question 9.
Analyse the three components of a political party. (2016 OD)
Answer:
The three components of a political party are as follows:

The leaders. A political party consists of leaders, who contest elections and if they win the elections, they perform the administrative jobs.
The active members. They are the ones who work actively for the party. They are the assistants of the leaders and implement the plans and ideologies of the political party.
The followers. They are the ardent followers of the parties and their leaders and support them in the elections.

Question 10.
Explain any five suggestions to reform political parties in India. (2011 D)
Or
Suggest and explain any five effective measures to reform political parties. (2016 OD, 2015 OD)
Answer:
Five suggestions made to reform the political parties:

Law to regulate the internal affairs of political parties like maintaining a register of its members, to follow its own constitution, to have independent authority, to act as judge in case of party dispute, to hold open elections to the highest post.
It should be mandatory for political parties to give one-third tickets to women candidates. Also there should be quota for women on the decision-making bodies of the party.
There should be state funding of elections. The government should give money to parties to support their election expenses in kind (petrol, paper, telephone, etc.) or in cash on the basis of votes secured by the party in the previous election.
The candidate should be educated, so that he can solve and understand people’s problems. His previous record should be cleared. He should be honest and there should be no criminal case against him.
Citizens can reform politics if they take part directly and join political parties. People can put pressure on political parties through petitions, publicity in media, agitations etc.

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Chapter 5 Popular Struggles and Movements Class 10 Important Questions | NCERT Social- Science Chapter-1 – Edu grown

Question 1.
Explain with examples the two types of political movements. (2011)
Answer:
The movement in Nepal and movement in Bolivia are examples of two types of political movements for democracy.
The movement in Nepal was to regain popular control over the government from the King. This was a struggle to restore democracy. The movement in Bolivia was against a specific policy of the elected democratic government. The people of Bolivia agitated and protested against the government’s decision of privatization of water.

Both these movements are instances of political conflicts that led to popular struggles. Even though in both cases public demonstration of mass support clinched the dispute, their level of impact was different.

Question 2.
What inspiration do we get from Bolivia’s popular struggle? Explain any three values that we can learn from it. (2014)
Answer:
We can identify the following values in Bolivia’s struggle:

It was a conflict between the people and the government to fight for justice and fairness and to fight against the greed of the government.
The Bolivian organization FEDECOR represented the common or general interest. The members of the organization did not necessarily benefit from the cause that the organization represented. They were fighting for collective social good.
The political conflict that led to popular struggle in Bolivia involved mass mobilization. It showed the power of the common people.

Question 3.
Mention any three similarities between struggles of Nepal and Bolivia. (2012)
Answer:
The struggle in both these countries relates to establishing and restoring democracy. The success of peoples’ struggle is a reminder that popular struggles are integral to the working of democracy. The democratic struggle in Nepal and Bolivia share some elements:

The popular struggle in the form of protest turned into indefinite strike.
Struggle involved mass mobilization.
Political conflict led to popular struggle.
Political organization played a critical role.

Question 4.
Explain with examples, how movements are different from interest groups. (2013)
Answer:
Movements:

Movements have a loose organization.
Movements are issue specific and long-term involving more than one issue.
Their decision-making is more informal and flexible.
They depend much on spontaneous mass participation.
Example: Narmada Bachao Andolan under Medha Patkar.

Interest groups:

Interest groups form organizations and undertake activities to promote their interests.
Interests groups are both sectional and public. Sectional interest groups promote interest of particular section of society and promotional or public interest groups aim to help groups other than their own members.
They promote collective good and are concerned with welfare of the society and not just their own members.
Example: BAMCEF (Backward and Minorities Community Employee Federation).

Question 5.
Describe the movement for democracy in Nepal. (2011)
Answer:
The Nepalese for democracy arose with the specific objective of reversing the king’s order that led to suspension of democracy. The popular struggle in Nepal involved many organizations other than political parties like the SPA or the Nepalese Communist Party. All the major labour unions and their federations joined the movement. Many other organizations of the indigenous people, teachers, lawyers and human rights groups extended support to the movement.

Question 6.
Differentiate between sectional interest groups and public interest groups with examples.
Answer:

Sectional interest groups	Public interest groups
(i) They seek to promote the interest of a particular section or group of society.	(i) They seek to promote collective interest rather than selective good.
(ii) They are sectional groups because they represent a section of society.	(ii) They are public groups because they represent the general people of the society.
(iii) They promote selective good and are concerned only about the interest of their section of the society, their own members and not the society in general.	(iii) They promote collective good and are concerned with welfare of the society and not just their own members.
(iv) For example, FEDECOR (Bolivian organization).	(iv) E.g., BAMCEF (Backward and Minorities Community Employee Federation).

Question 7.
Explain how the relationship between political parties and pressure groups can take different forms? (2011)
Answer:
The relationship between political parties and pressure groups can take different forms, some direct and others very indirect.
In some instances the pressure groups are either formed or led by the leaders of political parties or act as extended arms of political parties. For example, most trade unions and students’ organizations in India are either established by or affiliated to one or the other major political party.

Sometimes political parties grow out of movements. For example, the Assam Movement led by students against the ‘foreigners’ led to the formation of the Asom Gana Parishad. The roots of parties like the DMK and the AIADMK in Tamil Nadu can be traced to social reform movement during the 1930s and 1940s. When the relationship between parties and interest groups is not so direct they often take positions opposed to each other. Yet they are in dialogue and negotiation. New issues raised by movements have been taken up by political parties.

Question 8.
“The struggle of the Nepali people is a source of inspiration to democrats all over the world.” Support the statement. (2015)
Answer:

The Nepalese movement for democracy arose with the specific objective of reversing the king’s order that led to suspension of democracy.
The movement of 2006 was aimed at regaining popular control over the government from the king.
The popular struggle in Nepal involved many organisations other than political parties like the SPA or the Nepalese Communist Party.
All major political parties in the Parliament formed a Seven Party Alliance (SPA) and called a four day strike in Kathmandu. This strike turned into an indefinite strike in which the Maoists and other insurgent groups joined hands.
All the major labour unions and their federations joined the movement. Many other organisations of the indigenous people, teachers, lawyers and human rights groups extended support to the movement.
The movement put forward three demands:
- Restoration of Parliament
- Power to an all-party government
- A new Constituent Assembly.
The number of protesters reached between three to five lakhs. They stuck to their demands and the king was forced to concede to all three demands. On 24th April, the SPA chose Girija Prasad Koirala as the new Prime Minister of the interim government.

Question 9.
What are sectional interest groups? Describe their functioning. (2016)
Answer:
Sectional interest groups are the groups that seek to promote the interests of a particular section or a group of society. For example, FEDECOR (Bolivian organisation).
Functioning:

They perform a meaningful role in countering the undue influence of other groups.
They create awareness about the needs and concerns of their own society.

Question `10.
What is the difference between pressure group and a political party ?
Answer:

Pressure groups	Political parties
(i) Pressure groups are organisations that attempt to influence government policies.	(i) A political party is a group of people who come together to contest elections and hold powers in the government.
(ii) Unlike political parties, pressure groups do not aim to directly control or share political powers. The groups wield power without responsibility.	(ii) Political parties form and run governments. They play a decisive role in making laws, shaping public opinion.
(iii) Pressure groups are not accountable to the people.	(iii) Political parties have to face the people in elections. Parties have to be responsive to peoples’ need and demands. Otherwise people can reject them in next elections.

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Ch 4 Gender, Religion and Caste Class 10 Important Questions | NCERT Social- Science Chapter-4 – Edu grown –

Question 1.
Explain the status of women’s representation in India’s legislative bodies. (2014)
Answer:
The one way to ensure that women related problems get adequate attention is to have more women as elected representatives. To achieve this, it is legally binding to have a fair proportion of women in the elected bodies.

Panchayati Raj in India has reserved one-third seats in Local Government bodies for women.
In India, the proportion of women in legislature has been very low. The percentage of elected women members in Lok Sabha is not even 10 per cent and in State Assemblies less than 5 per cent. India is behind several developing countries of Africa and Latin America. Women organisations have been demanding reservations of at least one-third seats in Lok Sabha and State Assemblies for women.
And only recently, in March 2010, the women’s reservation bill was passed in the Rajya Sabha ensuring 33% reservation to women in Parliament and State Legislative bodies.

Question 2.
How does religion influence the political set up in our country? Explain. (2015)
Answer:
Gandhiji said, “Religion can never be separated from politics”. By religion he did not mean any particular religion like Hinduism or Islam, but moral values and ethics drawn from religion to guide politics. Religion in politics is not as dangerous as it may seem to us. Ethical values of each religion can play a major role in politics. According to human rights groups, most of the victims of communal riots in our country are from religious minorities.

Government can take special steps to protect them. Family laws of all religions discriminate against women. The government can change laws to make them more equitable. These instances show a relationship between religion and politics. People should be able to express in politics their needs, interests and demands as members of a religious community. Thus, it is the responsibility of those whose political power is able to regulate the practice of religion, to prevent discrimination and oppression. These political acts are not wrong as long as they treat every religion equally.

Question 3.
“Gender division is not based on Biology but on social expectations and stereotypes’. Support the statement. (2012)
Or
Mention different aspects of life in which women are discriminated against or disadvantaged in India.
Answer:
‘Gender division is not based on Biology but on social expectations and stereotypes’:

Boys and girls are brought up to believe that the main responsibility of women is house work and bringing up children. There is sexual division of labour in most families where women stay at home and men work outside to play the role of breadwinners.
Literacy rate among women is only 54% in comparison to 76% among men. In studies, girls mostly perform better than boys, but they drop out simply because parents prefer to spend their resources on their sons’ education. A smaller proportion of girls go for higher studies.
On an average, a woman works more than an average man everyday. Since much of her work is not paid for, therefore often not valued. The Equal Wages Act provides for equal wages for equal work, but in almost all areas of work from sports to cinema, from factories to fields, women are comparatively paid less because of the male chauvinistic bent of mind of society.
Child sex-ratio (number of girl children per thousand boys) is very low. In India, the national average is 927. In some places it is even lower because parents prefer to have sons so they get girl child aborted.
In urban areas too, women are not respected and are unsafe even in their homes being subjected to beating, harassment and other forms of domestic violence.
The role of women in politics in most societies is minimal.

Question 4.
State how caste inequalities are still continuing in India.
Answer:
Caste has not disappeared from contemporary India and caste division is special to India. Some of the older aspects of caste persist even today.

Even now most people marry within their own caste.
Untouchability has not ended completely despite constitutional prohibition.
Effects of centuries of advantages and disadvantages can be felt today. The caste groups that had access to education under old system have done well, whereas those groups that did not have access to education have lagged behind.
There is a large presence of ‘upper caste’ among the urban middle classes in our country.
Caste continues to be linked to economic status as is evident from National Sample Survey:
- The average economic status of caste groups still follows the old hierarchy
- the ‘upper’ castes are best off, the Dalits and Adivasis are worst off and the backward classes are in between.
- Although every caste has some poor members, the proportion of those living in extreme poverty is higher for lowest castes and much lower for the upper castes.
- Similarly, every caste has some members who are rich. The upper castes are heavily over-represented among the rich while the lower castes are under-represented.
- The SC, ST and OBC together account for about two-thirds of India’s population.

Question 5.
What factors have brought about a change in the Indian Caste system in modern times? Explain. (2012)
Answer:
The castes and caste system in modern India have undergone great changes due to the efforts of social reformers and the socio-economic changes in India. With economic development, large scale urbanisation, growth of literacy and education, occupational mobility and weakening of the position of landlords in the villages, the old notions of caste hierarchy are breaking down. Politics too influences the caste system and caste identities by bringing them into the political arena. The Constitution of India prohibits any caste-based discrimination and has laid foundations of policies to reverse injustices of the caste system.

Question 6.
Explain the various forms that caste can take in politics.
Answer:
Various forms of caste in politics:

When governments are formed, political parties usually take care that representatives of different castes and tribes find a place in it.
When parties choose candidates, they keep in mind the composition of the electorate and accordingly choose candidates from different castes so as to muster necessary support to win elections.
Political parties make appeals to caste sentiments to gain support. Some political parties are known to favour some castes.
Universal adult franchise and the principle of one-person-one-vote compelled political leaders to mobilise political support. It also brought new consciousness among people belonging to those castes which were treated as inferiors.

Question 7.
What was the Feminist Movement? Explain the political demands of the Feminist Movement in India. (2013)
Answer:
Feminist Movements are radical women’s movements aiming at attaining equality for women in personal and family life and public affairs. These movements have organised and agitated to raise channels for enhancing the political and legal status of women and improving their educational and career opportunities.

Political demands of the feminist movement in India:
The one way to ensure that women related problems get adequate attention in India is to have more women as elected representatives. To achieve this, it is legally binding to have a fair proportion of women in the elected bodies. Panchayati Raj in India has reserved one-third seats in Local Government bodies for women.

In India, the proportion of women in legislature has been very low. The percentage of elected women members in Lok Sabha is not even 10 per cent and in State Assemblies less than 5 per cent. India in behind several developing countries of Africa and Latin America. Women organizations have been demanding reservations of at least one-third seats in Lok Sabha and State Assemblies for women.

And only recently, in March 2010, the women’s reservation bill was passed in the Rajya Sabha ensuring 33 per cent reservation for women in Parliament and State Legislative bodies.

Question 8.
What have been the consequences of the political expression of gender division in free India? (2013)
Answer:
Political expression of gender division and political mobilisation has helped improve women’s role in public life all over the world including India. However, despite some improvements since Independence, ours is still a male-dominated society and women lag behind in all fields.

Literacy rate among women is only 54 per cent as compared with 76 per cent among men.
Proportion of women among highly paid and valuable jobs in still very small.
Equal Wages Act provides that equal wages should be paid for equal work. However, in all areas from sports and cinema, factories to fields, women are paid less than men for the same amount of work.
In many parts of India, parents prefer to have sons and find ways to abort the girl child before she is born.

Question 9.
Explain the factors that have led to the weakening of the caste system in India. (2014)
Answer:
Reasons which have contributed to changes in caste system:

Efforts of political leaders and social reformers like Gandhiji, B.R. Ambedkar who advocated and worked to establish a society in which caste inequalities are absent.
Socio-economic changes such as:
- urbanisation
- growth of literacy and education
- occupational mobility
- weakening of landlord’s position in the village
- breaking down of caste hierarchy have greatly contributed.
The Constitution of India prohibited any caste-based discrimination and laid the foundations of policies to reverse the injustices of the caste system.
Provision of fundamental rights has played a major role because these rights are provided to all citizens without any discrimination.

Question 10.
How far is it correct to say that it is not politics that gets caste ridden but it is the caste that gets politicised? Explain. (2015)
Answer:
Politics too influences the caste system and caste identities by bringing them into the political arena. This takes several forms:

Each caste group tries to become bigger by incorporating within itself neighbouring castes or sub¬castes.
Various caste groups enter into a coalition with other castes for negotiations.
New caste groups like ‘backward’ and ‘forward’ have come up in the political arena.
Expressions of caste differences in politics give many disadvantaged communities the chance to demand their share of power and thus gain access to decision-making.
Many political and non-political organisations have been demanding and agitating for an end to discrimination against particular castes for more dignity and more access to land, resources and opportunities.

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