Exercise 12.2 Exponents and Powers- NCERT Solutions Class 8

Go back to  'Exponents and Powers'

Question 1

Express the following numbers in standard form.

(i) \(0.0000000000085\)

(ii) \(0.00000000000942\)

(iii) \(6020000000000000\)

(iv) \(0.00000000837\)

(v) \(31860000000\)

Solution

Video 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 \(3.186 \times {10^{10}}\)

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) \(3.61492 \times 10^{6}\)

Solution

Video Solution

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

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

Video 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) \(0.07\,\text{mm}\)

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

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

Video 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

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

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school

0