To find the angular velocity of the rod when it hits the table, we can use the principle of conservation of energy. Initially, the rod is in a vertical position with its center of mass at a height of \(\frac{L}{2}\) from the pivot point. When it falls to the horizontal position, the center of mass is at a height of \(\frac{L}{3}\) from the pivot.

Step-by-step Solution:

First, calculate the initial potential energy (PE_i) of the rod:

\(PE_i = Mgh\), where \(h = \frac{L}{2}\).

So, \(PE_i = Mg\frac{L}{2}\).

At the horizontal position, calculate the final potential energy (PE_f).

The center of mass is at height \(\frac{L}{3}\):

\(PE_f = Mg\frac{L}{3}\).

Using conservation of energy:

Initial Potential Energy = Final Potential Energy + Rotational Kinetic Energy (KE_rot)

\(Mg\frac{L}{2} = Mg\frac{L}{3} + \frac{1}{2}I\omega^2\)

The moment of inertia \((I)\) of the rod about the pivot is:

\(I = \frac{1}{3}ML^2\) (using the parallel axis theorem).

Substitute \(I\) in the energy conservation equation:

\(Mg\frac{L}{2} - Mg\frac{L}{3} = \frac{1}{2} \times \frac{1}{3}ML^2 \omega^2\)

Simplify and solve for \(\omega\):

\(\frac{MgL}{6} = \frac{1}{6}ML^2\omega^2\)

Cancel \(M\):

\(gL = L^2\omega^2\)

\(\omega^2 = \frac{3g}{L}\)

Therefore, the angular velocity is:

\(\omega = \sqrt{\frac{3g}{L}}\).

Thus, the correct answer is \(\sqrt{\frac{3g}{L}}\).