A uniform disc of radius \(R\) and mass \(M\) is free to oscillate about the axis \(A\) as shown in the figure. For small oscillations the time period is ______.
(\(g\) is acceleration due to gravity)

From the given figure, the axis of rotation is a horizontal line passing through the point \(A\) on the edge of the disc. The circular arrow around the axis indicates that the disc oscillates out of its plane, meaning the axis of rotation is a tangent to the disc in its own plane.

The moment of inertia of a uniform disc about its diameter is \(I_{cm} = \dfrac{1}{4}MR^2\).

Using the parallel axis theorem, the moment of inertia of the disc about the given axis of rotation is:
\(I = I_{cm} + MR^2 = \dfrac{1}{4}MR^2 + MR^2 = \dfrac{5}{4}MR^2\)

The distance from the center of mass of the disc to the axis of rotation is \(d = R\).

The time period of a physical pendulum is given by:
\(T = 2\pi\sqrt{\dfrac{I}{Mgd}}\)

Substituting the values of \(I\) and \(d\):
\(T = 2\pi\sqrt{\dfrac{\dfrac{5}{4}MR^2}{MgR}} = 2\pi\sqrt{\dfrac{5R}{4g}}\)

Answer: \(2\pi\sqrt{\dfrac{5R}{4g}}\)