Correct option is c $\frac{1}{2 \pi} \sqrt{\frac{1}{L C}-\frac{R^{2}}{L^{2}}}$
Megritude of admittanos $Y=\frac{1}{Z}$ is given by:
$$
|\mathrm{Y}|=\frac{\mid \mathrm{H}^{2} \mathrm{X}^{2}+\left(\mathrm{H}^{2}+\mathrm{X}^{2}-\mathrm{X}_{1} \mathrm{X}, \mathrm{X}\right)^{\prime}}{\left.\mathrm{X}_{1} \mid \mathrm{R}^{2}+\mathrm{X}_{2}^{2}\right)} \frac{1}{2}
$$
Non, $|Z|$ is maximum or $|Y|$ is minimum at $\omega=$ m, such that
$$
\left[\mathrm{H}^{2}+\mathrm{X}_{\mathrm{L}}^{2}-\mathrm{X}_{\mathrm{L}_{\mathrm{L}}} \mathrm{X}_{\mathrm{c}}=0\right.
$$
whare $X_{L}=w_{t} L$ and $X_{c}=\frac{1}{w_{2} C}$ $O e R^{2}+\infty_{1} L^{2}=\omega_{1} L \times \frac{1}{\omega_{1} C}=\frac{L}{C}$
$\mathrm{or} \mathrm{\omega}^{2}=\frac{1}{\mathrm{LC}}-\frac{\mathrm{R}^{2}}{\mathrm{~L}^{2}}$
$\sigma r \omega,=\sqrt{\frac{1}{L C}-\frac{R^{2}}{L^{2}}}$