a
(a)

The time period of the disc is \(2 \pi \sqrt{(3 r / 2 g )}\)

We know that the time period of an object,

\(T =2 \pi \sqrt{(I / mgL )}\)

where,

\(I=\) moment of inertia from the suspended point

\(L =\) distance of its centre from suspended point \(= r\)

we know that, the moment of inertia of disc about its centre \(= mr ^2 / 2\)

using parallel axis theoram the moment of inertia from a point in its periphery,

\(I = mr ^2+ mr ^2 / 2=3 mr ^2 / 2\)

putting the values in the above equation we get,

\(2 \pi \sqrt{\left(3 mr ^2 / 2 mgr \right)}\)

\(=2 \pi \sqrt{(3 r / 2 g )}\)

therefore, the time period of the disc is \(2 \pi \sqrt{(3 r / 2 g )}\)