In figure E and v_ cm represent the total energy and speed of...

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 \quad \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$$\begin{aligned} \frac{3}{4} &=\frac{1}{2}\left[1+\frac{k^{2}}{R^{2}}\right] \\ \Rightarrow \quad \frac{3}{2}-1 &=\frac{k^{2}}{R^{2}} \\ \Rightarrow \quad \frac{k^{2}}{R^{2}} &=\frac{1}{2} \end{aligned}$$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.