Answer :
To determine which of the given options is an irrational number, we need to evaluate each expression under the square root and check if the result is a rational number or irrational number.
### Option (a) [tex]\(\sqrt{\frac{9}{49}}\)[/tex]
1. [tex]\(\frac{9}{49} = \frac{3^2}{7^2}\)[/tex]
2. [tex]\(\sqrt{\frac{9}{49}} = \sqrt{\frac{3^2}{7^2}} = \frac{3}{7}\)[/tex]
Since [tex]\(\frac{3}{7}\)[/tex] is a rational number, option (a) is rational.
### Option (b) [tex]\(\sqrt{\frac{4}{27}}\)[/tex]
1. [tex]\(\frac{4}{27} = \frac{2^2}{3^3}\)[/tex]
2. [tex]\(\sqrt{\frac{4}{27}} = \sqrt{\frac{2^2}{3^3}} = \frac{2}{\sqrt{27}}\)[/tex]
Since [tex]\(\sqrt{27} = 3\sqrt{3}\)[/tex], [tex]\(\frac{2}{\sqrt{27}} = \frac{2}{3\sqrt{3}}\)[/tex], which can be simplified but still involves an irrational component [tex]\(\sqrt{3}\)[/tex]. Thus, option (b) is irrational.
### Option (c) [tex]\(\sqrt{\frac{9}{25}}\)[/tex]
1. [tex]\(\frac{9}{25} = \frac{3^2}{5^2}\)[/tex]
2. [tex]\(\sqrt{\frac{9}{25}} = \sqrt{\frac{3^2}{5^2}} = \frac{3}{5}\)[/tex]
Since [tex]\(\frac{3}{5}\)[/tex] is a rational number, option (c) is rational.
### Option (d) [tex]\(\sqrt{\frac{2}{8}}\)[/tex]
1. [tex]\(\frac{2}{8} = \frac{1}{4}\)[/tex]
2. [tex]\(\sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\)[/tex]
Since [tex]\(\frac{1}{2}\)[/tex] is a rational number, option (d) is rational.
### Conclusion
After evaluating all options, we find that:
- Option (a) is rational.
- Option (b) is irrational.
- Option (c) is rational.
- Option (d) is rational.
Therefore, the irrational option is:
(b) [tex]\(\sqrt{\frac{4}{27}}\)[/tex]
### Option (a) [tex]\(\sqrt{\frac{9}{49}}\)[/tex]
1. [tex]\(\frac{9}{49} = \frac{3^2}{7^2}\)[/tex]
2. [tex]\(\sqrt{\frac{9}{49}} = \sqrt{\frac{3^2}{7^2}} = \frac{3}{7}\)[/tex]
Since [tex]\(\frac{3}{7}\)[/tex] is a rational number, option (a) is rational.
### Option (b) [tex]\(\sqrt{\frac{4}{27}}\)[/tex]
1. [tex]\(\frac{4}{27} = \frac{2^2}{3^3}\)[/tex]
2. [tex]\(\sqrt{\frac{4}{27}} = \sqrt{\frac{2^2}{3^3}} = \frac{2}{\sqrt{27}}\)[/tex]
Since [tex]\(\sqrt{27} = 3\sqrt{3}\)[/tex], [tex]\(\frac{2}{\sqrt{27}} = \frac{2}{3\sqrt{3}}\)[/tex], which can be simplified but still involves an irrational component [tex]\(\sqrt{3}\)[/tex]. Thus, option (b) is irrational.
### Option (c) [tex]\(\sqrt{\frac{9}{25}}\)[/tex]
1. [tex]\(\frac{9}{25} = \frac{3^2}{5^2}\)[/tex]
2. [tex]\(\sqrt{\frac{9}{25}} = \sqrt{\frac{3^2}{5^2}} = \frac{3}{5}\)[/tex]
Since [tex]\(\frac{3}{5}\)[/tex] is a rational number, option (c) is rational.
### Option (d) [tex]\(\sqrt{\frac{2}{8}}\)[/tex]
1. [tex]\(\frac{2}{8} = \frac{1}{4}\)[/tex]
2. [tex]\(\sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\)[/tex]
Since [tex]\(\frac{1}{2}\)[/tex] is a rational number, option (d) is rational.
### Conclusion
After evaluating all options, we find that:
- Option (a) is rational.
- Option (b) is irrational.
- Option (c) is rational.
- Option (d) is rational.
Therefore, the irrational option is:
(b) [tex]\(\sqrt{\frac{4}{27}}\)[/tex]