As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be

Let the mass m of the bob to fall through height h.
The charge in the bob's potential energy is transformed into the kinetic energies of the bob and the flywheel, according to the work-energy theorem.

A flywheel's rotational kinetic energy is provided by

$\frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{2}\frac{{\mathrm{mr}}^2}{2}{\mathrm{\omega }}^2=\frac{{\mathrm{mr}}^2{\mathrm{\omega }}^2}{4}$

Since the thread is wrapped around the flywheel, the relationship between

w of the flywheel according to the formula v=wr

$\mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{r}}^2+\frac{{\mathrm{mr}}^2{\mathrm{\omega }}^2}{4}$

$=\frac{3{\mathrm{m\omega }}^2{\mathrm{r}}^2}{4}$

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

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

$\mathrm{\omega }=\frac{1}{\mathrm{r}}\sqrt{\frac{4\mathrm{gh}}{3}}$

Hence the correct answer is $\frac{1}{\mathrm{r}}\sqrt{\frac{4\mathrm{gh}}{3}}.$