Which one of the following is the smallest surd?

\sqrt{4}

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\sqrt{27}

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\sqrt{9}

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\sqrt{5}

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Solution

The correct option is B $\sqrt{4}$
As we know, $4 < 5 < 9 < 27$
So, $\sqrt{4}$ < $\sqrt{5}$ < $\sqrt{9}$ < $\sqrt{27}$
Hence, the answer is $\sqrt{4}$ .

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