A thin uniform rod \(( X )\) of mass M and length L is pivoted at a height \(\left(\frac{ L }{3}\right)\) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is _________ . (g = gravitational acceleration)

Explanation:
To find the correct angular velocity of the rod when it hits the table top, we apply the law of conservation of mechanical energy step-by-step:

1. Determine the Center of Mass (CM) Position:
The center of mass of a thin uniform rod of length \( L \) lies at its geometric center, which is at a distance of \( \frac{L}{2} \) from either end.
According to the figure, the pivot is located at a distance of \( \frac{L}{3} \) from the bottom end. Therefore, the distance (\( d \)) between the pivot and the center of mass is:
\( d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6} \)

2. Calculate the Moment of Inertia (\( I \)) about the Pivot:
Using the parallel axis theorem (\( I = I_{\text{cm}} + Md^2 \)):
\( I = \frac{1}{12}ML^2 + M\left(\frac{L}{6}\right)^2 \)
\( I = \frac{1}{12}ML^2 + \frac{1}{36}ML^2 = \frac{3 + 1}{36}ML^2 = \frac{4}{36}ML^2 = \frac{1}{9}ML^2 \)

3. Calculate the Loss in Potential Energy (\( \Delta \text{PE} \)):
- In the initial vertical position, the center of mass is located at a height of \( \frac{L}{6} \) directly above the pivot point (table level).
- In the final horizontal position, the entire rod lies flat on the table, meaning its center of mass drops down to the exact level of the pivot.
Thus, the net vertical displacement of the center of mass is \( \Delta h = \frac{L}{6} \), giving a potential energy loss of:
\( \Delta \text{PE} = Mg\Delta h = Mg\left(\frac{L}{6}\right) \)

4. Apply Conservation of Energy:
The loss in potential energy is fully converted into rotational kinetic energy (\( \frac{1}{2}I\omega^2 \)):
\( \frac{1}{2}I\omega^2 = \Delta \text{PE} \)
\( \frac{1}{2}\left(\frac{1}{9}ML^2\right)\omega^2 = Mg\left(\frac{L}{6}\right) \)
\( \frac{1}{18}ML^2\omega^2 = \frac{1}{6}MgL \)

Simplifying the equation to solve for \( \omega \):
\( \omega^2 = \frac{18}{6}\frac{g}{L} = \frac{3g}{L} \)
\( \omega = \sqrt{\frac{3g}{L}} \)

The mathematical derivation confirms that the provided answer key is completely accurate.

Hence, Option D is the correct choice.