A uniform circular disc of radius R oscillates in a vertical plane about a horizontal axis, its minimum time period is $\mathrm{\pi }\sqrt{\frac{\mathrm{B}\sqrt{2\mathrm{R}}}{\mathrm{g}}}$. The value of B is _______.

The time period of a physical pendulum (or compound pendulum) is minimum when the distance between point of suspension and COM (i.e. l) is equal to the radius of gyration about its COM i.e. l = k.

The moment of inertia of a disc about an axis perpendicular to its plane and passing through its COM is

$\begin{array}{l}\mathrm{I}={\mathrm{mk}}^2=\frac{1}{2}{\mathrm{mR}}^2\Rightarrow \mathrm{k}=\frac{\mathrm{R}}{\sqrt{2}}\\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}\left({\mathrm{k}}^2+{\mathrm{l}}^2\right)}{\mathrm{mgl}}}\\ \Rightarrow {\mathrm{T}}_{min}=2\mathrm{\pi }\sqrt{\frac{\frac{{\mathrm{R}}^2}{2}+\frac{{\mathrm{R}}^2}{2}}{\mathrm{gR}/\sqrt{2}}}\Rightarrow {\mathrm{T}}_{min}=2\mathrm{\pi }\sqrt{\frac{\sqrt{2}\mathrm{R}}{\mathrm{g}}}=\mathrm{\pi }\sqrt{\frac{4\sqrt{2}\mathrm{R}}{\mathrm{g}}}\\ \Rightarrow \mathrm{B}=4\end{array}$