A disc of mass 1kg and radius R is free of rotate about a...

Let the angular speed of disc be $\omega$.

Using conservation of mechanical energy

$mg2R = \dfrac{1}{2}{I}_{disc}{\omega}^{2} + \dfrac{1}{2}{I}_{particle}{\omega}^{2}$

Where, $I$ is moment of inertia.

$⇒ mg2R = \dfrac{{\omega}^{2}}{2}\left[\dfrac{{m}^{2}}{2} + m{R}^{2}\right]$

$⇒ mg2R = \dfrac{{\omega}^{2}}{2}\left(\dfrac{3}{2}m{R}^{2}\right)$

$⇒ \dfrac{3}{4}{\omega}^{2} = \dfrac{2g}{R}$

$⇒ {\omega}^{2} = \dfrac{8g}{3R}$

Thus, angular speed is $\omega = \sqrt{\dfrac{80}{3R}}rad{s}^{- 1}$

Given $\omega = 4\sqrt{\dfrac{x}{3R}}$

Comparing both, $16\dfrac{x}{3R} = \dfrac{80}{3R}$

Therefore, the value of $x = 5$.