When alternating current is passed through L-R series circuit, the power factor is $\frac{\sqrt{3}}{2}$ and $R=50 \Omega$, then the value of $L$ is $\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

Given: Resistance $R = 50 \Omega$, power factor $\cos \phi = \frac{\sqrt{3}}{2}$.

The phase angle satisfies $\phi = \pm\frac{\pi}{6}$, taking $\phi = \frac{\pi}{6}$ as standard.

For an LR circuit, $\tan \phi = \frac{X_L}{R}$. Substituting known values:

$\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} = \frac{X_L}{50}$

Solving gives $X_L = \frac{50}{\sqrt{3}} \Omega$.

Using $X_L = \omega L$ with $\omega = 2\pi f$:

$L = \frac{50/\sqrt{3}}{2\pi f}$

Assuming a standard frequency of 50 Hz:

$L = \frac{50}{2\pi\sqrt{3} \cdot 50} = \frac{1}{2\pi\sqrt{3}}$ Henry

This corresponds to option C.