# Exercise 12.2 Exponents and Powers- NCERT Solutions Class 8

## Chapter 12 Ex.12.2 Question 1

Express the following numbers in standard form.

(i) $$0.0000000000085$$

(ii) $$0.00000000000942$$

(iii) $$6020000000000000$$

(iv) $$0.00000000837$$

(v) $$31860000000$$

### Solution

(i) $$0.0000000000085$$

What is known?

Usual form

What is unknown?

Standard form

Reasoning:

How to use –$$a.b \times {10^n}$$

Where a is a whole number, b is a decimal number and n is an integer. Small numbers are expressed with negative exponent i.e. n is negative integer.

Steps:

To convert this small number in to standard from we need to move decimal to its right by $$12$$ steps.

 $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ $$13$$ $$0$$ $$\bf{.}$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$8$$ $$5$$

\begin{align}&0.0000000000085 \\&= \frac{{85}}{{10000000000000}}\\&= \frac{{85}}{{10000000000000}} \\&= \frac{{8.5 \times 10}}{{{{10}^{13}}}}\\&= 8.5 \times {10^{ - 13}} \times {10^1}\\&= 8.5 \times {10^{ - 12}}\end{align}

$$\therefore$$ Standard form of given number is $$8.5 \times {10^{ - 12}}$$

(ii) $$0.00000000000942$$

What is known?

Usual form

What is unknown?

Standard form

Reasoning:

How to use – $$a.b \times {10^n}$$

Where a is a whole number, b is a decimal number and n is an integer. Small numbers are expressed with negative exponent i.e. n is negative integer.

Steps:

 $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ $$13$$ $$14$$ $$0$$ $$\bf{.}$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$9$$ $$4$$ $$2$$

\begin{align}&0.00000000000942 \\&= \frac{{942}}{{100000000000000}}\\&= \frac{{942}}{{100000000000000}}\\&= \frac{{9.42 \times {{10}^2}}}{{{{10}^{14}}}}\\&= 9.42 \times {10^{ - 14}} \times {10^2}\\&= 9.42 \times {10^{ - 12}}\end{align}

$$\therefore$$ Standard form of given number is $$9.42 \times {10^{ - 12}}$$

(iii) $$6020000000000000$$

What is known?

Usual form

What is unknown?

Standard form

In this question a big number has to be converted to its standard form. In this case $$n$$ is positive to represent this number as $$a \cdot b \times {10^{\rm{n}}}$$ and the decimal will be moved to its left to represent this number in its standard form as bellow.

 $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ $$13$$ $$14$$ $$15$$ $$16$$ $$6$$ $$0$$ $$2$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$

$6020000000000000 = 6.02 \times {10^{15}}$

Decimal has moved by $$15$$ steps so standard form of given number is $$6.2 \times {10^{15}}$$

(iv) $$0.00000000837$$

What is known?

Usual form

What is unknown?

Standard form

Reasoning:

How to use –$$a.b \times {10^n}$$

Where a is a whole number, b is a decimal number and $$n$$ is an integer. Small numbers are expressed with negative exponent i.e. $$n$$ is negative integer.

Steps:

 $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$0$$ $$\bf{.}$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$8$$ $$3$$ $$7$$

\begin{align}&0.000000000837 \\&= \frac{{837}}{{100000000000}}\\ &= \frac{{8.37 \times {{10}^2}}}{{{{10}^{11}}}}\\&= 8.37 \times {10^{ - 11}} \times {10^2}\\&= 8.37 \times {10^{ - 9}}\end{align}

$$\therefore$$ Standard form of given number is $$8.37 \times {10^{ - 9}}$$

(v) $$31860000000$$

What is known?

Usual form

What is unknown?

Standard form

Reasoning:

In this question a big number has to be converted to its standard form. In this case n is positive to represent this number as $$a \cdot b \times {10^{\rm{n}}}$$and the decimal will be moved to its left to represent this number in its standard form as below.

Steps:

 $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$3$$ $$1$$ $$8$$ $$6$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$

$31860000000 = 3.186 \times {10^{10}}$

So, decimal has moved by $$10$$ steps to its left

$$\therefore$$ Standard form of given number is $$3.186 \times {10^{10}}$$

## Chapter 12 Ex.12.2 Question 2

Express the following numbers in usual form.

(i) $$3.02 \times 10^{-6}$$

(ii) $$4.5 \times 10^{4}$$

(iii) \begin{align}\text{3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{-8}}\end{align}

(iv) $$1.0001 \times 10^{9}$$

(v) $$5.8 \times 10^{12}$$

(vi) $$3.61492 \times 10^ What is known? Standard form What is unknown? Usual form Reasoning: As we know standard form of any small or larger number is \(a.b \times {10^n}$$,Where $$a$$ is a whole number, $$b$$ is a decimal number and $$n$$ is an integer

For small number $$n$$ is negative and for large number $$n$$ is positive. So, to convert a small number to its usual form we need to move decimal to its left by number of steps given as exponent values.

Steps:

(i) $$\,3.02 \times {10^{ - 6}}$$

$$\therefore$$ Its usual form is

\begin{align}3.02 \times {10^{ - 6}} &= \frac{{3.02}}{{1000000}}\\&= 0.00000302\end{align}

(ii) $$\,4.5 \times {10^4}$$

Now to convert a big number to its usual form we need to move decimal to its right by number of steps given as its exponent.

$4.5 \times {10^4}= 45000$

$$\therefore$$ Answer is $$45000$$

(iii) $$\,3 \times {10^{ - 8}} = {{ }}0.00000003$$

(iv) $$1.0001 \times {10^9} = 1000100000$$

(v) $$\,5.8 \times {10^{12}} = {{ }}5800000000000$$

(vi) $$\,3.61492 \times {10^6} = 3614920$$

\)

{6}

## Chapter 12 Ex.12.2 Question 3

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

(i) $$1$$ micron is equal to \begin{align}\frac{1}{{100000}}\,\,{\rm{m}}\end{align}

(ii) Charge of an electron is $$0.000,000,000,000,000,000,16$$ coulomb.

(iii) Size of a bacteria is $$0.0000005\, \rm{m}$$

(iv) Size of a plant cell is $$0.00001275\, \rm{m}$$

(v) Thickness of a thick paper is $$0.07 \,\rm{mm}$$

### Solution

What is known?

Usual form

What is unknown?

Standard form

Reasoning:

How to use –$$a.b \times {10^n}$$

Where a is a whole number, b is a decimal number and n is an integer. Small numbers are expressed with negative exponent i.e. $$n$$ is negative integer.

Steps:

(i) $$1$$ micron is equal to \begin{align}\frac{1}{{1000000}}\; \rm{m}\end{align}

\begin{align}\frac{1}{{1000000}} & =0.000001\\ &=1\times10^{-6}\end{align}

(ii) $$0.00000000000000000016$$

\begin{align}& 0.00000000000000000016\\ &=\frac{16}{100000000000000000000} \\& =\frac{1.6\times 10}{{{10}^{20}}} \\& =1.6\times {{10}^{-19}}\,\,\text{coulomb}\end{align}

(iii) $$0.0000005 \;\rm{m}$$

\begin{align}0.0000005\text{m}&=\frac{5}{10000000} \\& =\frac{5}{{{10}^{7}}} \\& =5\times {{10}^{-7}}\text{m} \end{align}

(iv) $$0.00001275 \;\rm{m}$$

{\begin{align}0.00001275\,\text{m}~&=\frac{1275}{100000000} \\& =\frac{1.275\times {{10}^{3}}}{{{10}^{8}}} \\& =1.275\times {{10}^{-5}} \rm m\end{align}}

(v) $$0.07\,\text{mm}$$

\begin{align} 0.07\text{mm}&=\frac{7}{100} \\& =\frac{7}{{{10}^{2}}} \\& 7\times {{10}^{-2}}\text{mm} \end{align}

## Chapter 12 Ex.12.2 Question 4

In a stack there are $$5$$ books each of thickness $$20\, \rm{mm}$$ and $$5$$ paper sheets each of thickness $$0.016\,\rm{mm}$$.

What is the total thickness of the stack?

### Solution

What is known?

$$5$$ books of $$20\rm{mm}$$ thickness each. $$5$$ paper sheets of $$0.016\rm{mm}$$ thickness each.

What is unknown?

Total thickness of the stack

Reasoning:

First find the total thickness of $$5$$ books then find out total thickness of $$5$$ paper sheets. Now find total thickness of stack by adding these two.

Steps:

Thickness of each book $$= 20 \,\rm{mm}$$

Number of books per stack $$= 5$$

Thickness of $$5$$ books $$=5\times20=10\,\rm{mm}$$

Thickness of each sheet$$=0.016\,\rm{mm}$$

Thickness of 5 paper sheets

$$=5\times 0.016=0.080\,\,\text{mm}$$

Total thickness of stack

\begin{align} &= 100 +0.080 \\ & =100.080\;\rm{mm}\\ & =100.080 \\ & =1.0008\times10^2\,\rm{mm} \end{align}

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