Question

The square root of 841 is 29, than square root of 0.00000841 is equal to:
A.0.029
B.0.0029
C.0.00029
D.0.29

Answer 1

Hint: Write 0.00000841 as a product of 841 and ${{10}^{-8}}$ use “$\sqrt{ab}=\sqrt{a}\times \sqrt{b}$” to calculate the value of 0.00000841. Put $a=841$ and $b={{10}^{-8}}$ and further solve to get the answer.

Complete step-by-step answer:
Given the square root of 841 is 29. We know that,
Square root of a number is defined as the number which when multiplied by itself produces the original number. Square root of a number ‘a’ is denoted by “$\sqrt{a}$” .
According to the question, $\sqrt{841}=29$ And we have to find the square root of 0.00000841.
i.e. $\sqrt{0.00000841}=?$
By the properties of square root, we know that $\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}$ .
$0.00000841=841\times {{10}^{-8}}$
Let us take $a=841$ and $b={{10}^{-8}}$ in the above property of square root. Now we will get,
$\sqrt{0.00000841}=\sqrt{841}\times \sqrt{{{10}^{-8}}}$
$\sqrt{841}=29$ (Given in question)
$\sqrt{{{10}^{-8}}}={{10}^{-4}}$ (As ${{10}^{-4}}\times {{10}^{-4}}={{10}^{-8}}$ , see the definition of square root above).
Now, on putting $\sqrt{841}=29$ and $\sqrt{{{10}^{-8}}}={{10}^{-4}}$ , we will get
$\begin{align}
  & \sqrt{0.00000841}=29\times {{10}^{-4}} \\
 & =0.0029 \\
\end{align}$
Hence, the required value of $\sqrt{0.00000841}$ is 0.0029 and option (b) is the correct answer.

Note:In the solution we have used that $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$. We can prove this as follows.
Let us take $x=\sqrt{ab}$ and $y=\sqrt{a}\times \sqrt{b}$ .
On squaring x, ${{\left( x \right)}^{2}}={{\left( \sqrt{ab} \right)}^{2}}=ab$
On squaring y, ${{\left( y \right)}^{2}}={{\left( \sqrt{a}\times \sqrt{b} \right)}^{2}}$
$\begin{align}
  & \Rightarrow {{\left( y \right)}^{2}}={{\left( \sqrt{a} \right)}^{2}}\times {{\left( \sqrt{b} \right)}^{2}} \\
 & \Rightarrow {{\left( y \right)}^{2}}=a\times b \\
 & \Rightarrow {{\left( y \right)}^{2}}=ab \\
 & \Rightarrow \left( {{y}^{2}} \right)=ab \\
\end{align}$
So, ${{x}^{2}}={{y}^{2}}$
i.e. ${{\left( \sqrt{ab} \right)}^{2}}={{\left( \sqrt{a}\sqrt{b} \right)}^{2}}$
Taking all the terms to LHS, we will get,
$\Rightarrow {{\left( \sqrt{ab} \right)}^{2}}-{{\left( \sqrt{a}\sqrt{b} \right)}^{2}}=0$ .
We know \[\left( {{x}^{2}}-{{y}^{2}} \right)=\left( x-y \right)\left( x+y \right)\] , So
$\Rightarrow \left( \sqrt{ab}-\sqrt{a}\sqrt{b} \right)\left( \sqrt{ab}+\sqrt{a}\sqrt{b} \right)=0$
As $\sqrt{ab}+\sqrt{a}\sqrt{b}$ will be zero only when $a=0$ and $b=0$ . So for all cases,
$\left( \sqrt{ab}-\sqrt{a}\sqrt{b} \right)=0$
$\Rightarrow \sqrt{ab}=\sqrt{a}\sqrt{b}$.