a
Mass of the circular disc, \(m=10 \,kg\)

Radius of the disc, \(r=15 \,cm =0.15\, m\)

The torsional oscillations of the disc has a time period, \(T=1.5\, s\)

The moment of inertia of the disc is:

\({I}=\frac{1}{2} m r^{2}\)

\(={2} \times(10) \times(0.15)^{2}\)

\(=0.1125 \,kg\, m ^{2}\)

\(T=2 \pi \sqrt{\frac{I}{\alpha}} a\)

Time period,

is the torsional constant.

\(\alpha=\frac{4 \pi^{2} I}{T^{2}}\)

\(=\frac{4 \times(\pi)^{2} \times 0.1125}{(1.5)^{2}}\)

\(=1.972\, Nm / rad\)

Hence, the torsional spring constant of the wire is \(1.972\, Nm\) \(rad\) \(^{-1}\).