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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:

BasisEconomic ActivitiesNon-Economic Activities
1. MeaningThese 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/MotiveThese 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 SatisfactionThese 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 ExamplesPeople 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. LogicThese 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 DifferenceIndustryCommerceTrade
1. MeaningIt 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. CapitalIt requires a huge amount ’ of fixed and working capital for productionIt requires limited fixed capital but huge working capital.It requires small capital as per turnover.
3. ScopeIt covers genetic, manufacturing, and constructive industries.It covers trade and auxiliaries to trade.Does it include? ‘Mental and external trade
4. BasisIt is the basis of modern business.Its basis is industries and professions.it is based upon commerce.
5. UtilityGoods 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 workIndustries are established at a particular place. It may be a work-shop or factory.Here, goods are transferred from one place to other placesThe 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 |

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.

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.

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}

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

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.

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
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.4 4

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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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
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 2NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 3

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
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 5

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
NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.5 7

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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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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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
NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise 7

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

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

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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-3 g 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 γ2 in 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.

(c) Wavelength of light, λ ≈ 10-5 cm = 0.00001 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

(c) 0.2370 g cm-3
► 4

(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 × 10-10 m
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, Va = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3
Molar volume of 1 mole of hydrogen atoms at STP,
Vm = 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 × 1011 m. 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, Vb = 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, Vbe= 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 = VSodium
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 / VSodium
Substituting the value from above, we get
Density of sodium atom, ρ =4.67 × 10-3 Kg 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 A1/3
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 A1/3
r0 = 1.2 f = 1.2 × 10-15 m
Volume of nucleus, V = (4 / 3) π r3
= (4 / 3) π (r0 A1/3)3 = (4 / 3) π r0 A    ….. (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,
ρSodium = 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 / mpme2 c3G)

Where,
t = Age of Universe
e = Charge of electrons = 1.6 ×10–19 C
ε0 = Absolute permittivity
mp = Mass of protons = 1.67 × 10–27 kg
me = 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 × 109 Nm2/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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