## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers (EX 12.2) Exercise 12.2

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## Access NCERT Solution for Class 8 Maths Chapter 12- Exponents and Powers

Refer to pages 1 - 7 for exercise 12.2 in the PDF

EXERCISE 12.2

1. Express the following numbers in standard form.

(i) \[0.0000000000085\]

Ans: A number is in the standard form, if it is written in the form $a \times {10^b}$ where $a$ is any natural or decimal number and $b$ is any whole number.

As we move the decimal from left to right, then the power of 10 increases with a negative sign.

Therefore,

\[0.0000000000085\] can be written as,

\[{\text{0}}{\text{.0000000000085}} = 8.5 \times {10^{ - 12}}\]

(ii) \[0.00000000000942\]

Ans: A number is in the standard form, if it is written in the form $a \times {10^b}$ where $a$ is any natural or decimal number and $b$ is any whole number.

As we move the decimal from left to right, then the power of 10 increases with a negative sign.

Therefore,

\[0.00000000000942\] can be written as,

\[0.00000000000942 = 9.42 \times {10^{ - 12}}\]

(iii) \[6020000000000000\]

Ans: A number is in the standard form, if it is written in the form $a \times {10^b}$ where $a$ is any natural or decimal number and $b$ is any whole number.

As we move the decimal from right to left, then the power of 10 increases with a positive sign.

Therefore,

\[6020000000000000\] can be written as,

\[{\text{6020000000000000}} = 6.02 \times {10^{15}}\]

(iv) \[0.00000000837\]

As we move the decimal from left to right, then the power of 10 increases with a negative sign.

Therefore,

\[0.00000000837\] can be written as,

\[0.00000000837 = 8.37 \times {10^{ - 9}}\]

(v) \[31860000000\]

As we move the decimal from right to left, then the power of 10 increases with a positive sign.

Therefore,

\[31860000000\] can be written as,

\[31860000000 = 3.186 \times {10^{10}}\]

2. Express the following numbers in usual form.

(i) \[{\text{3}}{\text{.02}} \times {\text{1}}{{\text{0}}^{ - 6}}\]

Ans: A number is in the usual form, if it is not written in the form $a \times {10^b}$ where $a$ is any natural or decimal number and $b$ is any whole number and is simplified and written without the help of ${10^b}$.

In expanded form this can be written as,

\[{\text{3}}{\text{.02}} \times {\text{1}}{{\text{0}}^{ - 6}} = 0.00000302\]

(ii) \[4.5 \times {10^4}\]

Ans: A number is in the usual form, if it is not written in the form $a \times {10^b}$ where $a$ is any natural or decimal number and $b$ is any whole number and is simplified and written without the help of ${10^b}$.

In expanded form this can be written as,

\[4.5 \times {10^4} = 45000\]

(iii) \[3 \times {10^{ - 8}}\]

Ans: A number is in the usual form, if it is not written in the form $a \times {10^b}$ where $a$ is any natural or decimal number and $b$ is any whole number and is simplified and written without the help of ${10^b}$.

In expanded form this can be written as,

\[3 \times {10^{ - 8}} = 0.00000003\]

(iv) \[1.0001 \times {10^9}\]

In expanded form this can be written as,

\[1.0001 \times {10^9} = 1000100000\]

(v) \[5.8 \times {10^{12}}\]

In expanded form this can be written as,

\[5.8 \times {10^{12}} = 5800000000000\]

(vi) \[3.61492 \times {10^6}\]

In expanded form this can be written as,

\[3.61492 \times {10^6} = 3614920\]

3. Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to \[\dfrac{1}{{1000000}}m\].

\[\dfrac{1}{{1000000}}m = 0.000001m\]

As we move the decimal from left to right then the power of 10 increases with a negative sign.

Therefore,

\[0.000001m\] can be written as,

\[0.000001m = 1 \times {10^{ - 6}}\].

(ii) Charge of an electron is \[0.000,000,000,000,000,000,16{\text{ }}coulomb\].

Charge on an electron is \[0.000,000,000,000,000,000,16{\text{ }}coulomb\] .

As we move the decimal from left to right then the power of 10 increases with a negative sign.

Therefore,

Charge on electron can be written as,

\[0.000,000,000,000,000,000,16{\text{ }}coulomb = 1.6 \times {10^{ - 19}}coulombs\].

(iii) Size of a bacteria is \[0.0000005m\].

Size of a bacteria is \[0.0000005m\].

As we move the decimal from left to right then the power of 10 increases with a negative sign.

Therefore,

Size of a bacteria can be written as,

\[0.0000005m = 5 \times {10^{ - 7}}m\].

(iv) Size of a plant cell is \[0.00001275m\].

Size of a plant cell is \[0.00001275m\].

As we move the decimal from left to right then the power of 10 increases with a negative sign.

Therefore,

Size of a plant cell can be written as,

\[0.00001275m = 1.275 \times {10^{ - 5}}\].

(v) Thickness of a thick paper is \[0.07mm\].

Thickness of a thick paper is \[0.07mm\].

As we move the decimal from left to right then the power of 10 increases with a negative sign.

Therefore,

Thickness of a thick paper can be written as,

\[0.07mm = 7 \times {10^{ - 2}}\] .

4. In a stack there are 5 books each of thickness \[20mm\] and 5 paper sheets each of thickness \[0.016mm\]. What is the total thickness of the stack?

Ans: There are a total of 5 books.

Thickness of each book is \[20mm\].

Therefore, total thickness of book can be calculated as,

\[ {\text{Total thickness of 5 books}}=5 \times 20mm \]

\[ {\text{Total thickness of 5 books}}= 100mm \]

There are a total of 5 paper sheets.

Thickness of each paper sheet is \[0.016mm\].

Therefore, total thickness of paper sheets can be calculated as,

Total thickness of 5 paper sheets \[ = 5 \times 0.016mm\]

\[ = 0.08mm\]

So, the total thickness of stack

= Thickness of 5 books + Thickness of 5 paper sheets

Thickness of 5 books + Thickness of 5 paper sheets = (100 + 0.08)mm

= 100.08mm

= 1.0008 × 10^{2}mm

Thus, the total thickness of the stack is 1.0008 × 10^{2}mm.

## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.2

Opting for the NCERT solutions for Ex 12.2 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.2 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 12 Exercise 12.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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