A bob of mass ' m ' is tied by a massless string whose other end is wound on a flywheel (disc) of radius ' $R$ ' and mass ' $m$ '. When released from the rest, the bob starts falling vertically downwards. If the bob has covered a vertical distance ' $h$ ', then angular speed of wheel will be (There is no slipping between string and wheel, $g$ - acceleration due to gravity)

Conservation of mechanical energy is applied to the system, taking the initial state when the bob is at rest and the final state after it has fallen distance $h$.

Initial potential energy is $mgh$, with zero kinetic energy for both bob and flywheel. Final energy consists of translational kinetic energy $\frac{1}{2}mv^2$ for the bob and rotational kinetic energy $\frac{1}{2}I\omega^2$ for the flywheel.

For a solid disc, $I = \frac{1}{2}mR^2$. The no-slip condition gives $v = R\omega$.

Equating initial and final energies:

$mgh = \frac{1}{2}m(R\omega)^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\omega^2$

Simplifying yields:

$mgh = \frac{3}{4}mR^2\omega^2$

Solving for $\omega$:

$\omega^2 = \frac{4gh}{3R^2}$

$\omega = \frac{2}{R}\sqrt{\frac{gh}{3}}$

This corresponds to option A.