A fly wheel having mass 3 kg and radius 5 m is free to rotate about...

Let the mass of the flywheel be $M = 3$ kg and its radius be $R = 5$ m.
Let the hanging mass be $m = 3$ kg and the distance descended be $h = 3$ m.
The flywheel is a disk, so its moment of inertia is $I = \frac{1}{2}MR^2$.
By conservation of energy, the loss in potential energy of the hanging mass equals the gain in kinetic energy of the system.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the string does not slip, $v = R\omega$, so $\omega = \frac{v}{R}$.
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2$
Substituting the values: $3 \times 10 \times 3 = \frac{1}{2}(3)v^2 + \frac{1}{4}(3)v^2$
$90 = \frac{3}{2}v^2 + \frac{3}{4}v^2 = \frac{9}{4}v^2$
$v^2 = \frac{90 \times 4}{9} = 40$ m$^2$/s$^2$.
The kinetic energy of the wheel is $K_{wheel} = \frac{1}{2}I\omega^2 = \frac{1}{4}Mv^2$.
$K_{wheel} = \frac{1}{4} \times 3 \times 40 = 30$ J.