(C) $\sqrt{2}$ is a constant polynomial.$A$ constant polynomial is a polynomial of degree $0$.We can express $\sqrt{2}$ as $\sqrt{2} \cdot x^0$,where

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(C) $\sqrt{2}$ is a constant polynomial.$A$ constant polynomial is a polynomial of degree $0$.We can express $\sqrt{2}$ as $\sqrt{2} \cdot x^0$,where the exponent of the variable $x$ is $0$.Therefore,the degree of the polynomial $\sqrt{2}$ is $0$.Hence,$(c)$ is the correct answer.