Sol.

$$
I_{\text {disc }}=\frac{m R^2}{4}+m R^2=\frac{5 m R^2}{4}
$$

∴ For small displacement

$$
\tau=-\operatorname{mgR} \theta
$$


$$
\therefore \alpha=-\frac{\mathrm{mgR}}{\frac{5 \mathrm{mR}^2}{4 \mathrm{~g}}} \theta
$$


$$
\therefore \omega=\sqrt{\frac{4 \mathrm{~g}}{5 \mathrm{R}}}
$$
$\therefore \mathrm{T}=2 \pi \sqrt{\frac{5 \mathrm{R}}{4 \mathrm{~g}}}$