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