NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

# NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

Edited By Ramraj Saini | Updated on Sep 10, 2023 06:53 PM IST

NCERT solutions for Class 10 Maths chapter 1 Real Numbers are discussed here. In Mathematics, there are two kinds of numbers - Real number and Imaginary number. Real numbers are those numbers that can be represented on the number line. NCERT solutions for Class 10 Maths Chapter 1 Real Numbers will help the students to prepare for the exams in a better way. The problems and solutions are prepared by expert teachers keeping in mind the Latest updated CBSE syllabus 2023. Students can practice these NCERT solutions for class 10 to get good hold on the concepts.

In this Class 10 Maths Chapter 1, we are going to talk about real numbers. The real numbers class 10 carry detailed explanations for each and every question in simple, comprehensive, and step by step explanation. NCERT solutions for other subjects and classes can be downloaded by clicking on the above link.

Also, see-

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers - Important Formulae

Types of Numbers:

• Natural Numbers: N = {1, 2, 3, 4, 5, ...}
• Whole Numbers: W = {0, 1, 2, 3, 4, 5, ...}
• Integers: {..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ...}
• Positive Integers: Z+ = {1, 2, 3, 4, 5, ...}
• Negative Integers: Z– = {-1, -2, -3, -4, -5, ...}
##### TOEFL ® Registrations 2024

Accepted by more than 11,000 universities in over 150 countries worldwide

##### PTE Exam 2024 Registrations

Register now for PTE & Save 5% on English Proficiency Tests with ApplyShop Gift Cards

Rational and Irrational Numbers:

• Rational Numbers: p/q where p and q are integers and q is not zero. (e.g., 3/7)
• Irrational Numbers: Cannot be expressed as p/q (e.g., π, √5)

Real Numbers: Numbers located on the number line, used in real-world applications.

Euclid's Division Algorithm (Lemma):

• For positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r ≤ b.
• (a is the dividend, b is the divisor, q is the quotient, and r is the remainder.)
##### JEE Main high scoring chapters and topics

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Fundamental Theorem of Arithmetic:

• Composite Numbers can be expressed as the product of prime numbers.

HCF and LCM by Prime Factorization:

• Highest Common Factor (HCF): Product of smallest power of common factors in numbers.
• Least Common Multiple (LCM): Product of greatest power of prime factors in numbers.

HCF (a, b) × LCM (a, b) = a × b

Free download NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers PDF for CBSE Exam.

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Intext Questions and Exercise)

Real Numbers class 10 Exercise: 1.1

225 > 135. Applying Euclid's Division algorithm we get

$225=135\times 1+90$

since remainder $\neq$ 0 we again apply the algorithm

$135=90\times 1+45$

since remainder $\neq$ 0 we again apply the algorithm

$90=45\times 2$

since remainder = 0 we conclude the HCF of 135 and 225 is 45.

38220 > 196. Applying Euclid's Division algorithm we get

$38220=196\times 195+0$

since remainder = 0 we conclude the HCF of 38220 and 196 is 196.

867 > 225. Applying Euclid's Division algorithm we get

$867=255\times 3+102$

since remainder $\neq$ 0 we apply the algorithm again.

since 255 > 102

$255=102\times 2+51$

since remainder $\neq$ 0 we apply the algorithm again.

since 102 > 51

$102=51\times 2+0$

since remainder = 0 we conclude the HCF of 867 and 255 is 51.

Let p be any positive integer. It can be expressed as

p = 6q + r

where $q\geq 0$ and $0\leq r< 6$

but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.

The maximum number of columns in which they can march = HCF (32, 616)

Since 616 > 32, applying Euclid's Division Algorithm we have

$616=32\times 19+8$

Since remainder $\neq$ 0 we again apply Euclid's Division Algorithm

Since 32 > 8

$32=8\times 4+0$

Since remainder = 0 we conclude, 8 is the HCF of 616 and 32.

The maximum number of columns in which they can march is 8.

[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1]

Let x be any positive integer.

It can be written in the form 3q + r where $q\geq 0$ and r = 0, 1 or 2

Case 1:

For r = 0 we have

x 2 = (3q) 2

x 2 = 9q 2

x 2 = 3(3q 2 )

x 2 = 3m

Case 2:

For r = 1 we have

x 2 = (3q+1) 2

x 2 = 9q 2 + 6q +1

x 2 = 3(3q 2 + 2q) + 1

x 2 = 3m + 1

Case 3:

For r = 2 we have

x 2 = (3q+2) 2

x 2 = 9q 2 + 12q +4

x 2 = 3(3q 2 + 4q + 1) + 1

x 2 = 3m + 1

Hence proved.

Let x be any positive integer.

It can be written in the form 3q + r where $q\geq 0$ and r = 0, 1 or 2

Case 1:

For r = 0 we have

x 3 = (3q) 3

x 3 = 27q 3

x 3 = 9(3q 3 )

x 3 = 9m

Case 2:

For r = 1 we have

x 3 = (3q+1) 3

x 3 = 27q 3 + 27q 2 + 9q + 1

x 3 = 9(3q 3 + 3q 2 +q) + 1

x 3 = 3m + 1

Case 3:

For r = 2 we have

x 3 = (3q+2) 3

x 3 = 27q 3 + 54q 2 + 36q + 8

x 3 = 9(3q 3 + 6q 2 +4q) + 8

x 3 = 3m + 8

Hence proved.

Real Numbers class 10 Exercise: 1.2

The number can be as a product of its prime factors as follows

$\\140=2\times 2\times 5\times 7\\ 140=2^{2}\times 5\times 7$

The given number can be expressed as follows

$\\156=2\times 2\times 3\times 13\\ 156=2^{2}\times 3\times 13$

The number is expressed as the product of the prime factors as follows

$\\3825=3\times 3\times 5\times 5\times 17\\ 3825=3^{2}\times 5^{2}\times 17$

The given number can be expressed as the product of its prime factors as follows.

$5005=5\times 7\times 11\times 13$

The given number can be expressed as the product of their prime factors as follows

$7429=17\times 19\times 23$

26 = 2 x 13

91 = 7 x 13

HCF(26,91) = 13

LCM(26,91) = 2 x 7 x 13 = 182

HCF x LCM = 13 x 182 = 2366

26 x 91 = 2366

26 x 91 = HCF x LCM

Hence Verified

The number can be expressed as the product of prime factors as

510 = 2 x 3 x 5 x 17

92 = 2 2 x 23

HCF(510,92) = 2

LCM(510,92) = 2 2 x 3 x 5 x 17 x 23 = 23460

HCF x LCM = 2 x 23460 = 46920

510 x 92 = 46920

510 x 92 = HCF x LCM

Hence Verified

336 is expressed as the product of its prime factor as

336 = 2 4 x 3 x 7

54 is expressed as the product of its prime factor as

54 = 2 x 3 3

HCF(336,54) = 2 x 3 = 6

LCM(336,54) = 2 4 x 3 3 x 7 = 3024

HCF x LCM = 6 x 3024 = 18144

336 x 54 = 18144

336 x 54 = HCF x LCM

Hence Verified

The numbers can be written as the product of their prime factors as follows

12 = 2 2 x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 2 2 x 3 x 5 x 7 = 420

The given numbers are written as the product of their prime factors as follows

17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 17 x 23 x 29 = 11339

The given numbers are written as the product of their prime factors as follows

8 = 2 3

9 = 3 2

25 = 5 2

HCF = 1

LCM = 2 3 x 3 2 x 5 2 = 1800

As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have

HCF (306, 657) x LCM (306, 657) = 306 x 657

$\\ LCM (306, 657) = \frac{306\times 657}{HCF (306, 657) }\\ LCM (306, 657) = \frac{306\times 657}{9}\\ LCM (306, 657) =22338$

By prime factorizing we have

6 n = 2 n x 3 n

A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6 n we can conclude that for no value of n 6 n will end with the digit 0.

7 x 11 x 13 + 13

= (7 x 11 + 1) x 13

= 78 x 13

= 2 x 3 x 13 2

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

= (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5

= 5 x 1008

After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.

The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.

Time taken by Sonia = 18 = 2 x 3 2

Time taken by Ravi = 12 = 2 2 x 3

LCM(18,12) = 2 2 x 3 2 = 36

Therefore they would again meet at the starting point after 36 minutes.

Class 10 Maths Chapter 1 Solutions Exercise: 1.3

Let us assume $\sqrt{5}$ is rational.

It means $\sqrt{5}$ can be written in the form $\frac{p}{q}$ where p and q are co-primes and $q\neq 0$

$\\\sqrt{5}=\frac{p}{q}$

Squaring both sides we obtain

$\\\left ( \sqrt{5} \right )^{2}=\left (\frac{p}{q} \right )^{2}\\ 5=\frac{p^{2}}{q^{2}}\\ p^{2}=5q^{2}$

From the above equation, we can see that p 2 is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number. $(i)$

Therefore p can be written as 5r

p = 5r

p 2 = (5r) 2

5q 2 = 25r 2

q 2 = 5r 2

From the above equation, we can see that q 2 is divisible by 5, Therefore q will also be divisible by 5 as 5 is a prime number. $(ii)$

From (i) and (ii) we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that $\sqrt{5}$ is rational was wrong. Hence proved that $\sqrt{5}$ is irrational.

Let us assume $3 + 2 \sqrt 5$ is rational.

This means $3 + 2 \sqrt 5$ can be wriiten in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\3+2\sqrt{5}=\frac{p}{q}\\ 2\sqrt{5}=\frac{p}{q}-3\\ \sqrt{5}=\frac{p-3q}{2q}\\$

As p and q are integers $\frac{p-3q}{2q}\\$ would be rational, this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $3 + 2 \sqrt 5$ is rational was wrong. Therefore $3 + 2 \sqrt 5$ is irrational.

(i) $1/ \sqrt 2$

Let us assume $\frac{1}{\sqrt{2}}$ is rational.

This means $\frac{1}{\sqrt{2}}$ can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\\frac{1}{\sqrt{2}}=\frac{p}{q}\\ \sqrt{2}=\frac{q}{p}$

Since p and q are co-prime integers $\frac{q}{p}$ will be rational, this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $\frac{1}{\sqrt{2}}$ is rational was wrong. Therefore $\frac{1}{\sqrt{2}}$ is irrational.

(ii) $7 \sqrt 5$

Let us assume $7 \sqrt 5$ is rational.

This means $7 \sqrt 5$ can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\7\sqrt{5}=\frac{p}{q}\\ \sqrt{5}=\frac{p}{7q}$

As p and q are integers $\frac{p}{7q}\\$ would be rational, this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $7 \sqrt 5$ is rational was wrong. Therefore $7 \sqrt 5$ is irrational.

Let us assume $6 + \sqrt 2$ is rational.

This means $6 + \sqrt 2$ can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\6+\sqrt{2}=\frac{p}{q}\\ \sqrt{2}=\frac{p}{q}-6\\ \sqrt{2}=\frac{p-6q}{q}$

As p and q are integers $\frac{p-6q}{q}$ would be rational, this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $6 + \sqrt 2$ is rational was wrong. Therefore $6 + \sqrt 2$ is irrational.

Class 10 Maths Chapter 1 Solutions Exercise: 1.4

$\frac{13}{3125}=\frac{13}{5^{5}}$

The denominator is of the form 2 a x 5 b where a = 0 and b = 5. Therefore the given rational number will have a terminating decimal expansion.

$\frac{17}{8}=\frac{17}{2^{3}}$

The denominator is of the form 2 a x 5 b where a = 3 and b = 0. Therefore the given rational number will have a terminating decimal expansion.

$\frac{64}{455}=\frac{64}{5\times 7\times 13}$

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

$\\\frac{15}{1600}=\frac{3\times 5}{2^{6}\times 5^{2}}\\ \frac{15}{1600}=\frac{3}{2^{6}\times 5}$

The denominator is of the form 2 a x 5 b where a = 6 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

$\frac{29}{343}=\frac{29}{7^{3}}$

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

$\frac{23}{2 ^3 \times 5 ^ 2 }$

The denominator is of the form 2 a x 5 b where a = 3 and b = 2. Therefore the given rational number will have a terminating decimal expansion.

$\frac{129 }{2 ^ 2\times 7^5 \times 5 ^ 7 }$

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

$\\\frac{6}{15}=\frac{2\times 3}{3\times 5}\\ \frac{6}{15}=\frac{2}{5}$

The denominator is of the form 2 a x 5 b where a = 0 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

$\\\frac{35}{50}=\frac{5\times 7}{2\times 5^{2}}\\\frac{35}{50}=\frac{7}{2\times 5}$

The denominator is of the form 2 a x 5 b where a = 1 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

$\\\frac{77}{210}=\frac{7\times 11}{2\times 3\times5\times 7 }\\ \frac{77}{210}=\frac{11}{2\times 3\times5}$

The denominator is not of the form 2 a x 5 b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

decimal expansions of rational numbers are

(i) $\frac{13}{3125}=0.00416$

(ii)&nbsnbsp; $\frac{17}{8}=2.125$

(iv) $\frac{15}{1600}=0.009375$

(vI) $\frac{23}{2^{3}\times 5^{2}}=\frac{23}{200}=0.115$

(viii) $\frac{6}{15}=\frac{2}{5}=0.4$

(ix) $\frac{35}{50}=\frac{7}{2\times 5}=0.7$

$\\43.123456789=\frac{43123456789}{10^{9}}\\ 43.123456789=\frac{43123456789}{2^{9}\times 5^{9}}$

The denominator is of the form 2 a x 5 b where a = 9 and b = 9. Therefore the given number is rational and has a terminating decimal expansion.

Since the decimal part of the given number is non-terminating and non-repeating we can conclude that the given number is irrational and cannot be written in the form $\frac{p}{q}$ where p and q are integers.

$43. \overline{123456789}$

As the decimal part of the given number is non-terminating and repeating, the number is rational but its denominator will have factors other than 2 and 5.

## Summary Of NCERT solutions for Real Numbers Class 10 Maths

The following points have been examined by us:

1. Euclid's Division Lemma states that for any positive integers a and b, there exist whole numbers q and r such that a = bq + r where 0 ≤ r < b.
2. Euclid's Division Algorithm, which is based on Euclid's division lemma, outlines the procedure for finding the highest common factor (HCF) of two positive integers a and b with a > b as follows: Step 1: Apply the division lemma to find q and r where a = bq + r, 0 ≤ r < b. Step 2: If r = 0, then the HCF is b. If r ≠ 0, apply the Euclid Lemma to b and r. Step 3: Continue the process until the remainder is zero. The divisor at this stage will be the HCF (a, b). Moreover, HCF (a, b) = HCF (b, r).
3. The Fundamental Theorem of Arithmetic affirms that any composite number can be represented (factorized) as a product of primes, and this factorization is unique, except for the order in which the prime factors appear.

## Real Numbers Class 10 Solutions - Chapter Wise

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

If interested students can also check exercises here:

### How to Use NCERT Solutions for Maths Class 10 Chapter 1 Real Numbers?

• Now you have gone through the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers and have a good knowledge of answer writing in a structural manner. It's time to practice various kinds of problems based on real numbers.

• After the completion of the NCERT syllabus, you can check past 5-year papers of board exams. Previous year papers will increase your dealing ability with a variety of questions.

• NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERT and the previous year papers, you can jump to the next chapters.

## NCERT Exemplar solutions - Subject wise

### Benefits of NCERT Solutions for Maths Class 10 Chapter 1 Real Numbers

• Using the NCERT Class 10 Maths solutions chapter 1, students will be able to confirm the right answers once they are done solving the questions themselves.

• Class 10 Maths Chapter 1 NCERT solutions are solved by the subject matter experts. Therefore, the answers to all the questions are reliable.

• NCERT Class 10 solutions for Chapter 1 Maths gives the step by step explanations to all the questions which makes it easy for the students to understand.

1. What are the main topics covered in the NCERT Solutions for Class 10 Maths Chapter 1?

NCERT Solutions for Class 10 Maths Real Numbers includes topics like Euclid's Division Algorithm, Euclid's Division Algorithm to Find the Highest Common Factor,  Prime & Composite Numbers, and Rational & Irrational Numbers. NCERT Solutions for ncert class 10 maths chapter 1  is designed by our experts to provide students with a solid foundation in the fundamental concepts and techniques of mathematics. it will help students to score well on tests as well as in board exams.

2. Are NCERT Solutions for Class 10 Maths Chapter 1 imprtant?

Yes, NCERT Solutions for maths class 10 chapter 1 are important for several reasons. Firstly, these solutions provide detailed explanations of the various concepts and principles covered in the chapter.

Secondly, the NCERT Solutions for chapter 1 maths class 10  often include a large number of examples and practice problems, which can help students apply their knowledge to solve real-world problems.

Thirdly, the NCERT Solutions for class 10 chapter 1 maths are typically written by experienced teachers or subject matter experts.

Overall, real numbers class 10 solutions are essential resources for students who want excel in maths.

3. What is the weightage of the chapter Real Numbers for CBSE board exam?

As per the CBSE, the total weightage of class 10 maths ch 1  Real Number is 4 marks in the final board exam. To get a good score follow the NCERT syllabus and problems from the NCERT book and the previous year's papers. Interested students can use class 10 maths chapter 1 solutions to command the concepts.

4. Where can I find the complete solutions of NCERT Class 10 Maths?

The NCERT solutions for all 15 chapters of Class 10 Mathematics can be downloaded from the careers360 website. these solutions are listed chapter-wise in tabulate format above in this article. students can visit maths ch 1 class 10 solutions if they are facing any issues while solving the maths ch 1 class 10 problems.

5. How many chapters are there in the Class 10 Maths?

There are 15 chapters in Class 10 Maths NCERT syllabus. it includes Real Numbers, Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions, Triangles, Coordinate Geometry, Introduction to Trigonometry, Some Applications of Trigonometry, Circles, Constructions, Area Related to Circles, Surface Areas and Volumes, Statistics, and Probability.

## Upcoming School Exams

Admit Card Date:01 June,2024 - 20 June,2024

#### Odisha Council of Higher Secondary Education 12th Examination

Others:10 June,2024 - 20 June,2024

#### Nagaland Board High School Leaving Certificate Examination

Exam Date:11 June,2024 - 21 June,2024

#### Nagaland Board of Secondary Education 12th Examination

Exam Date:11 June,2024 - 21 June,2024

#### Kerala Department of Higher Secondary Education 12th Examination

Exam Date:12 June,2024 - 20 June,2024

Edx
1111 courses
Coursera
792 courses
Udemy
327 courses
Futurelearn
142 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD
Lancaster University, Lancaster
Bailrigg, Lancaster LA1 4YW
Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9