If $\bar{a}$ and $\overline{\mathrm{b}}$ are unit vectors such that $|\bar{a}+\overline{\mathrm{b}}|=\sqrt{3}$, then the angle between $\bar{a}$ and $\overline{\mathrm{b}}$ is

Given that $\bar{a}$ and $\bar{b}$ are unit vectors, $|\bar{a}| = |\bar{b}| = 1$. The magnitude of their sum is provided as $|\bar{a} + \bar{b}| = \sqrt{3}$. The magnitude squared is expanded as $|\bar{a} + \bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + 2 \bar{a} \cdot \bar{b}$, leading to $(\sqrt{3})^2 = 1^2 + 1^2 + 2 \bar{a} \cdot \bar{b}$.

Simplifying, $3 = 2 + 2 \bar{a} \cdot \bar{b}$, so $\bar{a} \cdot \bar{b} = \frac{1}{2}$. Using the dot product definition, $\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos \theta = \cos \theta$. Thus, $\cos \theta = \frac{1}{2}$, yielding $\theta = rccos(\frac{1}{2}) = \frac{\pi}{3}$.

Final Answer: $\boxed{\text{B}}$