The correct option of this question will be (a).

**Solution —**

Let $\theta$ be the angle made by $\sqrt{3} \hat{i}+\hat{j}$ with $x$-axis.

If $\hat{i}$ is the unit vector along $x$-axis, then

$\cos \theta=\frac{(\sqrt{3} \hat{i}+\hat{j}) \cdot \hat{i}}{|\sqrt{3} \hat{i}+\hat{j}||\hat{i}|}$

=$\frac{\sqrt{3}}{\sqrt{\sqrt{3}^{2}+1^{2}} \sqrt{1^{2}}}$

=$\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}$

or

$\theta=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=30^{\circ}$