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NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

Edited By Ramraj Saini | Updated on Mar 01, 2024 06:38 PM IST

Exponents and Powers Class 8 Questions And Answers provided here. These NCERT Solutions are created by expert team at craeers360 keeping the latest syllabus and pattern of CBSE 2023-23. In this article, you will get NCERT solutions for Class 8 Maths chapter 12 Exponents and Powers explained in a detailed manner to get a better understanding of the chapter. Also, you will learn how to make the representation of bigger numbers easier using about exponents and powers.

This Story also Contains
  1. Exponents and Powers Class 8 Questions And Answers PDF Free Download
  2. Exponents and Powers Class 8 Solutions - Important Formulae
  3. Exponents and Powers Class 8 NCERT Solutions (Intext Questions and Exercise)
  4. NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers - Topics
  5. NCERT Solutions for Class 8 Maths - Chapter Wise
  6. Key Features Of NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers
  7. NCERT Solutions for Class 8 - Subject Wise
  8. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers
NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

There are 11 questions in 2 exercises of the textbook. All these questions are explained in NCERT solutions for Class 8 Maths chapter 12 Exponents and Powers. Here you will get the detailed NCERT Solutions for Class 8 maths by clicking on the link.

Exponents and Powers Class 8 Questions And Answers PDF Free Download

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Exponents and Powers Class 8 Solutions - Important Formulae

  • Law of Product: am Ă— an = a(m + n)

  • Law of Quotient: am / an = a(m - n)

  • Law of Zero Exponent: a0 = 1

  • Law of Negative Exponent: a(-m) = 1 / am

  • Law of Power of a Power: (am)n = a(m * n)

  • Law of Power of a Product: (ab)n = an * bn

Law of Power of a Quotient: (a / b)m = am / bm

Free download NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers for CBSE Exam.

Exponents and Powers Class 8 NCERT Solutions (Intext Questions and Exercise)

Class 8 exponents and powers NCERT solutions - Topic 12.2 Powers With Negative Exponents

Question:(i) Find the multiplicative inverse of the following.

2^{-4}

Answer:

The detailed explanation for the question is written below,

The multiplicative inverse is \frac{1}{a^{m}}

So, the multiplicative inverse of 2^{-4} is 2^{4}

Question:(ii) Find the multiplicative inverse of the following.

10^{-5}

Answer:

Here is the detailed solution for the above question,

As we know,

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, the multiplicative inverse of 10^{-5} is 10^{5}

Question:(iii) Find the multiplicative inverse of the following.

7^{-2}

Answer:

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, the multiplicative inverse of 7^{-2} is 7^{2}

Question:(iv) Find the multiplicative inverse of the following.

5^{-3}

Answer:

we know,

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, for 5^{-3} multiplicative inverse is 5^{3}

Question:(v) Find the multiplicative inverse of the following.

10^{-100}

Answer:

The multiplicative inverse of a^{m} is \frac{1}{a^{m}}

So, the multiplicative inverse of 10^{-100} is 10^{100}

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Topic 12.2 Powers With Negative Exponents

Question:(i) Expand the following numbers using exponents.

1025.63

Answer:

1025.63 = 1\times 10^{3}+o\times 10^{2}+2\times 10^{1}+5\times 10^{0}+6\times 10^{-1}+3\times 10^{-2}

Question:(ii) Expand the following numbers using exponents.

1256.249

Answer:

1256.249

= 1\times 10^{3}+2\times 10^{2}+5\times 10^{1}+6\times 10^{0}+2\times 10^{-1}+4\times 10^{-2}+9\times 10^{-3}

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers - Topic 12.3 Laws Of Exponents

Question:1(i) Simplify and write in exponential form.

(-2)^{-3}\times (-2)^{-4}

Answer:

this is simplified as follows

(-2)^{-3}\times (-2)^{-4}

= \frac{1}{(-2)^{3}}\times \frac{1}{(-2)^{4}}

= \frac{1}{(-2)^{3+4}}=\frac{1}{(-2)^{7}}

= (-2)^{-7}

Question:1(ii) Simplify and write in exponential form .

p^3\times p^{-10}

Answer:

this is simplified as follows

p^3\times p^{-10}

p^{3-10} ............. [a^{m}\times a^{n}=a^{m+n}]

=p^{-7}

Question:1(iii) Simplify and write in exponential form.

3^2 \times 3^{-4}\times 3^6

Answer:

this can be simplified as follows

3^2 \times 3^{-4}\times 3^6

= 3^{2+(-4)+6} ............. [a^{m}\times a^{n}\times a^{o}=a^{m+n+o}]

= 3^{4}= 81

Class 8 maths chapter 12 question answer - Exercise: 12.1

Question:1 (i) Evaluate.

3^{-2}

Answer:

The detailed explanation for the above-written question is as follows,

We know that,

a^{-m}=\frac{1}{a^{m}}

So, here m =2 and a = 3

3^{-2}=\frac{1}{3^{2}} = \frac{1}{3}\times \frac{1}{3} = \frac{1}{9}

Question: 1(ii) Evaluate.

(-4)^{-2}

Answer:

The detailed explanation for the above-written question is as follows

We know that,

a^{-m}= \frac{1}{a^{m}}

So, here (a = -4) and (m = 2)

Then according to the law of exponent

(-4)^{-2}= \frac{1}{(-4)^{2}} = \frac{1}{(-4)}\times \frac{1}{(-4)} = \frac{1}{16} [ negative \times negative = positive]

Question: 1(iii) Evaluate.

\left(\frac{1}{2}\right )^{-5}

Answer:

The detailed solution for the above-written question is as follows

We know that,

(\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}} \& a^{-m} = \frac{1}{a^{m}}

So, here

a = 1 and b = 2 and m =-5

According to the law of exponent

(\frac{1}{2})^{-5}=\frac{1^{-5}}{2^{-5}} = \frac{-1}{2^{-5}}

=(-1)\times -2\times -2\times -2\times -2\times -2 = 32


Question: 2(i) Simplify and express the result in power notation with a positive exponent.

(-4)^{5}\div (-4)^{8}

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

\frac{a^{m}}{a_{n}} = a^{m-n} and a^{-m}= \frac{1}{a^{m}}

So according to this

a = -4, m = 5 and n = 8

\frac{4^{5}}{4_{8}} = -4^{5-8} = -4^{-3}

= (-\frac{1}{4})\times -(\frac{1}{4})\times(-\frac{1}{4}) = -\frac{1}{64}

Question: 2(ii) Simplify and express the result in power notation with positive exponent.

\left (\frac{1}{2^3} \right )^2

Answer:

The detailed solution for the above-written question is as follows

We know the exponential formula

\frac{a^{m}}{b^{m}} = (\frac{a}{b})^{m} and a^{-m}= \frac{1}{a^{m}} and (a^{m})^{n} = a^{mn}


So, we have given

a = 1, b=2

By using above exponential law,

\frac{a^{m}}{b^{m}} = \frac{1}{(2^{3})^{2}} = \frac{1}{2^{6}}

= \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} = \frac{1}{64}


Question: 2(iii) Simplify and express the result in power notation with a positive exponent.

(-3)^4\times \left(\frac{5}{3} \right )^{4}

Answer:

The detailed solution for the above-written question is as follows,

We know the exponential formula

(\frac{a}{b})^{m}= \frac{a^{m}}{b^{m}}

So, (-3)^{4}\times (\frac{5}{3})^{4}= (-3)^{4}\times \frac{5^{4}}{3^{4}} = \frac{625}{1}

Question: 2(iv) Simplify and express the result in power notation with a positive exponent.

(3^{-7}\div 3^{-10})\times 3^{-5}

Answer:

The detailed explanation for the above-written question is as follows

As we know the exponential form

\frac{a^{m}}{b^{n}}= a^{m-n} \& (a^{m}\times a^{n})=a^{m+n}

By using these two form we get,

\frac{3^{-7}}{3^{-10}}\times (3)^{-5}=3^{-7-(-10)} \times (3)^{-5}

=3^{3} \times (3)^{-5}= 3^{-2}

=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}

Question:2(v) Simplify and express the result in power notation with positive exponent.

2^{-3} \times (-7)^{-3}

Answer:

The detailed solution for the above-written question is as follows,

we know the exponential forms

a^{-m}=\frac{1}{a^{m}} & a^{m}\times b^{m}= (ab)^{m}

So, according to our data,

here initially we use first forms and then the second one.

2^{-3}\times (-7)^{-3}= \frac{1}{2^{3}}\times \frac{1}{(-7)^{3}}

= \frac{1}{(2\times -7)^{3}}= \frac{1}{(-14)^{3}}

Question:3(i) Find the value of.

(3^0 + 4^{-1})\times 2^2

Answer:

The detailed explanation for the above-written question is as follows,

As we know that a^{0}=1

So, 3^{0}=1

now,

=(1+\frac{1}{4})\times 2^{2}

=\frac{5}{4}\times 2^{2}\Rightarrow \frac{5}{2^{2}}\times 2^{2}=5/1

Question: 3(ii) Find the value of.

(2^{-1}\times 4^{-1})\div 2^{-2}

Answer:

The detailed explanation for the above-written question is as follows

Rewrite the equation

(2^{-1}\times 4^{-1})\div 2^{-2} = (2^{-1}\times 2^{-2})\div 2^{-2}

=(2^{-1+(-2)})\div 2^{-2} ................................. a^{m}\times a^{n}= a^{(m+n)}

=(2^{-3})\div 2^{-2} = 2^{-3-(-2)} ........................ a^{m}\div a^{n}= a^{(m-n)}

= 2^{-1}= \frac{1}{2}

Question: 3(iii) Find the value of.

\left (\frac{1}{2}\right)^{-2} + \left (\frac{1}{3}\right)^{-2} + \left (\frac{1}{4}\right)^{-2}

Answer:

The detailed explanation for the above-written question is as follows,

This is the exponential form

(a/b)^{m}= \frac{a^{m}}{b^{m}}

So, \left (\frac{1}{2}\right)^{-2} + \left (\frac{1}{3}\right)^{-2} + \left (\frac{1}{4}\right)^{-2}

=\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\frac{1}{4^{-2}} .......................using this form a^{m}= \frac{1}{a^{m}}

=2^{2}+3^{2}+4^{2}

= 4+9+16

= 29

Question: 3(iv) Find the value of.

(3^{-1} + 4^{-1} + 5^{-1})^0

Answer:

since we know that

a^{0}=1

(3^{-1} + 4^{-1} + 5^{-1})^0 =1

Question: 3(v) Find the value of.

\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}

Answer:

The detailed explanation for the above-written question is as follows

\left \{ \left (\frac{-2}{3} \right )^{-2} \right \}^{2}

=(-2/3)^{-2\times 2} .............. BY using these form of exponential (a^{m})^{n}=a^{mn}

(-2/3)^{-4}=(-3/2)^{4} ......... use this a^{-m}=\frac{1}{a^{m}}

=\frac{81}{16}

Question:4(i) Evaluate

\frac{8^{-1}\times 5^{3}}{2^{-4}}

Answer:

The detailed explanation for the above written question is as follows

\frac{8^{-1}\times 5^{3}}{2^{-4}}

after rewriting the above equation we get,

\frac{2^{-3}\times 5^{3}}{2^{-4}} =2^{-3-(-4)}\times 5^{3} ...........as we know that \frac{a^{m}}{a^{n}}= a^{m-n}

\\=2^{1}\times 5^{3}\\ = 2\times 125 = 250

An alternate method,

= \frac{5^{3}}{2^{-4}\times 2^{3}}

here you can use first a^{-m}= \frac{1}{a^{m}} and after that use a^{m}\times a^{n}= a^{m+n}

Question: 4(ii) Evaluate

(5^{-1}\times 2^{-1})\times 6^{-1}

Answer:

The detailed explanation for the above-written question is as follows

We clearly see that this is in the form of a^{-m}=\frac{1}{a^{m}}

So, (5^{-1}\times 2^{-1})\times 6^{-1}= (\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}

= \frac{1}{60}

Question: 5 Find the value of m for which

5^m \div 5^{-3} = 5^5

Answer:

We have,

a^{m}\div a^{n}= a^{m-n}

Here a = 5 and n =-3 and m-n = 5

therefore,

5^{m}\div 5^{-3}= 5^{m+3} = 5^{5}

By comparing from both sides we get

m+3 = 5

m= 2

Question: 6(i) Evaluate

\left \{\left (\frac{1}{3} \right )^{-1} - \left( \frac{1}{4} \right )^{-1} \right \}^{-1}

Answer:

The detailed solution for the above-written question is as follows

\left \{\left (\frac{1}{3} \right )^{-1} - \left( \frac{1}{4} \right )^{-1} \right \}^{-1}

= [(1\times 3)-(1\times 4)]^{-1} .............by using a^{-m}= \frac{1}{a^{m}}

= [3-4]^{-1}

\\= [-1]^{-1}\\=-1

Question: 6(ii) Evaluate

\left ( \frac{5}{8} \right )^{-7}\times \left (\frac{8}{5}\right)^{-4}

Answer:

The detailed solution for the above-written question is as follows

\left ( \frac{5}{8} \right )^{-7}\times \left (\frac{8}{5}\right)^{-4}

=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}} ................ using the form \frac{a^{m}}{b^{m}}= (a/b)^{m}

=5^{-7+4}\times 8^{-4+7} ............using a^{m} \div a^{n} = a^{m-n}

= 5^{-3}\times 8^{+3}

\\= \frac{8^{3}}{5^{3}}\\\\=\frac{512}{125}

Question: 7(i) Simplify

\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\;\;(t\neq 0)

Answer:

The detailed solution for the above-written question is as follows

\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\;\;(t\neq 0)

we can write 25= 5^{2}

So, after rewriting the equation,

\frac{5^{2}\times t^{-4}}{5^{-3}\times 10\times t^{-8}}

= \frac{5^{2+3}\times t^{-4+8}}{10} .................using the form [a^{m}\div a^{n}= a^{m-n}]

= \frac{5^{5}\times t^{4}}{10} .............(By expanding we have now)

= \frac{625t^{4}}{2}

Question: 7(ii) Simplify.

\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}

Answer:

The detailed solution for the above-written question is

\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times6 ^{-5}}

we can write 125 = 5^{3} and 6^{-5} can be written as (2\times 3)^{-5}

Now, rewriting the equation, we get

=\frac{3^{-5}\times 10^{-5}\times 5^{3}}{5^{-7}\times(2\times 3) ^{-5}}

=\frac{3^{-5}\times 10^{-5}\times 5^{3+7}}{(2\times 3) ^{-5}} .............by using [a^{m}\div a^{n}=a^{m-n}]

=\frac{ 10^{-5}\times 5^{10}}{(2) ^{-5}} .....................Use [a^{m}\div a^{n}=a^{m-n}]

5^{10-5}= 5^{5}=3125 .........................As [10^{-5} = (2\times 5)^{-5}=2^{-5}\times 5^{-5}] . 2^{-5} can be cancelled out with the denominator 2^{-5}


Class 8 maths chapter 12 NCERT solutions - Topic 12.4 Use Of Exponents To Express Small Numbers In Standard Form

Question: 1(i) Write the following numbers in standard form.

0.000000564

Answer:

the standard form of 0.000000564 is

\frac{564}{1000000000}=5.64\times10^{-7}

Question: 1(ii) Write the following numbers in standard form.

0.0000021

Answer:

The standard form 0.0000021 is

=\frac{21}{10000000}=2.1\times 10^{-6}

Question: 1(iii) Write the following numbers in standard form.

21600000

Answer:

The standard form 21600000 is

=2.16\times 10^{7}

Question: 1(iv) Write the following numbers in standard form.

15240000

Answer:

The standard form 15240000

=1.524\times 10^{7}

Question: 2 Write all the facts given in the standard form .

Answer:

  1. Distance between sun and earth 1.496\times10^{11}m

  2. speed of light is 3\times10^{8}m/s

  3. The avg. diameter of red blood cells is (7\times10^{-6}mm)

  4. the distance of the moon from the earth is (3.84467\times10^{8}m)

  5. size of the plant cell is (1.275\times10^{-5}m)

  6. The diameter of the wire on a computer chip is (3\times10^{-6}m)

  7. the height of the Mount Everest is (8.848\times10^{3}m)

NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers - Exercise: 12.2

Question: 1(i) Express the following numbers in standard form .

0.0000000000085

Answer:

The standard form is 8.5\times 10^{-12}

Question: 1(ii) Express the following numbers in standard form.

0.00000000000942

Answer:

The standard form is 9.42\times 10^{-12}

Question: 1(iii) Express the following numbers in standard form.

6020000000000000

Answer:

The standard form is 6.02\times 10^{15}

Question: 1(iv) Express the following numbers in standard form.

0.00000000837

Answer:

The standard form of the given number is 8.37\times10^{-9}

Question: 1(v) Express the following numbers in standard form.

31860000000

Answer:

The standard form is 3.186\times10^{10}

Question: 2(i) Express the following numbers in usual form.

3.02\times 10^{-6}

Answer:

3.02\times 10^{-6}

=\frac{3.02}{1000000}

=0.00000302

this is the usual form

Question: 2(ii) Express the following numbers in usual form.

4.5 \times 10^4

Answer:

4.5 \times 10^4=4.5\times 10000

=45000

this is the usual form

Question:2(iii) Express the following numbers in the usual form.

3\times 10^{-8}

Answer:

3\times 10^{-8}

\frac{3}{100000000} = 0.000000030

this is the usual form

Question: 2(iv) Express the following numbers in usual form.

1.0001\times 10^9

Answer:

1.0001\times 10^9

=1.0001\times 100000000

=1000100000

this is the usual form

Question: 2(v) Express the following numbers in usual form.

5.8\times 10^{12}

Answer:

5.8\times 10^{12}

=5.8\times 100000000000

=5800000000000

this is the usual form

Question:2(vi) Express the following numbers in usual form.

3.61492 \times 10^6

Answer:

3.61492 \times 10^6

= 3.61492\times 1000000

= 3614920

17155

Question:3(i) Express the number appearing in the following statements in standard form.

1 micron is equal to \frac{1}{1000000}m .

Answer:

1 micron is equal to

\frac{1}{1000000}m

= 1\times 10^{-6}

Question:3(ii) Express the number appearing in the following statements in standard form.

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

Answer:

Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.

=1.6\times 10^{-19} coulomb.

Question:3(iii) Express the number appearing in the following statements in standard form.

Size of a bacteria is 0.0000005 m.

Answer:

Size of a bacteria is 0.0000005 m

\frac{5}{10000000}=5\times 10^{-7}m

Question:3(iv) Express the number appearing in the following statements in standard form.

Size of a plant cell is 0.00001275 m.

Answer:

Size of a plant cell is0.00001275m

=\frac{1275}{10000}=1.275\times10^{-5}m

Question:3(v) Express the number appearing in the following statements in standard form.

Thickness of a thick paper is 0.07 mm

Answer:

The thickness of a thick paper is 0.07

=\frac{7}{100}=7\times 10^{-2}mm

Question:4 In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.

Answer:

the thickness of each book = 20mm

So, the thickness of 5 books = (5\times 20)=100 mm

the thickness of one paper sheet =0.016mm

So, the thickness of 5 paper sheet = (5\times 0.016)=0.08 mm

the total thickness of the stack = (100+0.08)mm

=100.08 mm or

(1.008\times10^{-2}mm)

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers - Topics

  • Powers with Negative Exponents
  • Laws of Exponents
  • Use of Exponents to Express Small Numbers in Standard Form
  • Comparing very large and very small numbers

NCERT Solutions for Class 8 Maths - Chapter Wise

Chapter -1

Rational Numbers

Chapter -2

Linear Equations in One Variable

Chapter-3

Understanding Quadrilaterals

Chapter-4

Practical Geometry

Chapter-5

Data Handling

Chapter-6

Squares and Square Roots

Chapter-7

Cubes and Cube Roots

Chapter-8

Comparing Quantities

Chapter-9

Algebraic Expressions and Identities

Chapter-10

Visualizing Solid Shapes

Chapter-11

Mensuration

Chapter-12

Exponents and Powers

Chapter-13

Direct and Inverse Proportions

Chapter-14

Factorization

Chapter-15

Introduction to Graphs

Chapter-16

Playing with Numbers

Key Features Of NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

Comprehensive Coverage: Solutions of maths chapter 12 class 8 cover all topics and concepts related to exponents and powers as per the Class 8 syllabus.

Step-by-Step Solutions: Detailed, step-by-step explanations for each problem, making it easy for students to understand and apply mathematical concepts related to exponents.

Variety of Problems: A wide range of problems, including exercises and additional questions, to help students practice and test their understanding of exponent rules, laws, and operations.

NCERT Solutions for Class 8 - Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. What are the important topics of chapter Exponents and Powers ?

The power of negative exponents, law of exponents and powers, and applications of power and exponents are the important topics of this chapter.

2. How does the NCERT solutions are helpful ?

NCERT solutions are helpful for the students if they are not able to NCERT problems on their own. These solutions are provided in a very detailed manner which will give them conceptual clarity.  

3. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

4. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

5. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

6. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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Mole fraction.
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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
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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
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