1. Interpret the setup

A uniform bar of length $L=1\,\text{m}$ is pivoted at one end on a horizontal table.

It is released from rest when it is at $30^\circ$ above the horizontal, and it falls until it becomes horizontal (hits the table).

We need the angular speed $\omega$ just before it hits the table.

2. Use conservation of mechanical energy

Since the bar rotates about a fixed pivot, we use:

$\text{Loss in gravitational potential energy} = \text{gain in rotational kinetic energy}$

3. Find the drop in height of the center of mass

For a uniform rod, the center of mass is at its midpoint, i.e. at distance

$\frac{L}{2} = \frac{1}{2}\,\text{m}$

from the pivot.

Initially, the rod makes angle $30^\circ$ with the horizontal, so the height of its center of mass above the table is

$h_i = \frac{L}{2}\sin 30^\circ = \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}\,\text{m}$

Finally, when it hits the table, the rod is horizontal, so the center of mass is at height

$h_f = 0$

Thus the vertical drop is

$\Delta h = \frac{1}{4}\,\text{m}$

So the loss in potential energy is

$mg\Delta h = mg\cdot \frac{1}{4}$

4. Moment of inertia of the rod about the pivot

For a uniform rod about one end:

$I = \frac{1}{3}mL^2$

Since $L=1$ m,

$I = \frac{1}{3}m$

5. Rotational kinetic energy at the instant it hits the table

$K = \frac{1}{2}I\omega^2 = \frac{1}{2}\cdot \frac{1}{3}mL^2\omega^2$

With $L=1$,

$K = \frac{1}{6}m\omega^2$

6. Apply energy conservation

$mg\cdot \frac{1}{4} = \frac{1}{6}m\omega^2$

Cancel $m$:

$\frac{g}{4} = \frac{\omega^2}{6}$

$\omega^2 = \frac{6g}{4} = \frac{3g}{2}$

Using $g=10\,\text{m/s}^2$ (standard JEE convention),

$\omega^2 = \frac{3\times 10}{2} = 15$

Hence,

$\omega = \sqrt{15}\,\text{s}^{-1}$

So,

$n=15$

7. Comparison with stored answer

Stored correct answer: $15$

Our derived answer: $15$

They match.