In figure \(E\) and \(v_{\mathrm{cm}}\) represent the total energy and speed of centre of mass of an object of mass \(1 \mathrm{~kg}\) in pure rolling. The object is

As we know, kinetic energy of an object in pure rolling motion is given as
\(E=\frac{1}{2} m v_{\mathrm{cm}}^{2}\left(1+\frac{k^{2}}{R^{2}}\right)\)
where, \(k\) is radius of gyration and \(m\) is the mass of the object.
\(\Rightarrow \frac{E}{v_{\mathrm{cm}}^{2}}=\frac{1}{2}\left[1+\frac{k^{2}}{R^{2}}\right] ...(i)\) \([\because\) Given, \(m=1 \mathrm{~kg}\) ]
From the given graph, substituting the value of \(\frac{E}{v_{\mathrm{cm}}^{2}}\) in Eq. (i), we get
\(\frac{3}{4} =\frac{1}{2}\left[1+\frac{k^{2}}{R^{2}}\right] \)
\( \Rightarrow \frac{3}{2}-1 =\frac{k^{2}}{R^{2}} \)
\( \Rightarrow \frac{k^{2}}{R^{2}} =\frac{1}{2}\)
Since, for a
(a) Sphere, \(\frac{k^{2}}{R^{2}}=\frac{2}{5}\)
(b) Ring, \(\frac{k^{2}}{R^{2}}=1\)
(c) Hollow cylinder, \(\frac{k^{2}}{R^{2}}=1\)
(d) Disc, \(\frac{k^{2}}{R^{2}}=\frac{1}{2}\)
So, the given object is disc.