A uniform disc of radius r is suspended through a small hole...

The time period of a physical pendulum is given by $T = 2\pi \sqrt{\dfrac{I}{mgd}}$.

For a uniform disc of radius $r$ and mass $m$, the moment of inertia about its center of mass is $I_{cm} = \dfrac{1}{2}mr^2$.

By the parallel axis theorem, the moment of inertia about a pivot at a distance $d$ from the center is $I = I_{cm} + md^2 = \dfrac{1}{2}mr^2 + md^2$.

Substituting $I$ into the time period formula, we get $T = 2\pi \sqrt{\dfrac{\dfrac{1}{2}mr^2 + md^2}{mgd}} = 2\pi \sqrt{\dfrac{\dfrac{r^2}{2d} + d}{g}}$.

To find the minimum time period, the expression $\left(\dfrac{r^2}{2d} + d\right)$ must be minimized.

Differentiating with respect to $d$ and equating to zero:

$\dfrac{d}{dd}\left(\dfrac{r^2}{2d} + d\right) = -\dfrac{r^2}{2d^2} + 1 = 0 \Rightarrow d = \dfrac{r}{\sqrt{2}}$.

Substituting $d = \dfrac{r}{\sqrt{2}}$ into the time period equation gives the minimum time period:

T_\min = 2\pi \sqrt{\dfrac{\dfrac{r^2}{2(r/\sqrt{2})} + \dfrac{r}{\sqrt{2}}}{g}} = 2\pi \sqrt{\dfrac{\dfrac{r}{\sqrt{2}} + \dfrac{r}{\sqrt{2}}}{g}} = 2\pi \sqrt{\dfrac{r\sqrt{2}}{g}}.