To solve this problem, we need to calculate the work done to bring each object to rest.$\\$The work done is equal to the change in kinetic energy, which is given by:$\\ W = \frac{1}{2} I \omega^2 \\$where $I$ is the moment of inertia and $\omega$ is the angular speed.$\\$Let's calculate the moment of inertia for each object:$\\$• For a solid sphere (A):$\\ I_A = \frac{2}{5} M R^2 \\$• For a thin circular disk (B):$\\ I_B = \frac{1}{2} M R^2 \\$• For a circular ring (C):$\\ I_C = M R^2 \\$Now, calculate the work done to bring each object to rest:$\\$• Work done for the solid sphere (A):$\\ W_A = \frac{1}{2} I_A \omega^2 = \frac{1}{2} \left(\frac{2}{5} M R^2\right) \omega^2 = \frac{1}{5} M R^2 \omega^2 \\$• Work done for the thin circular disk (B):$\\ W_B = \frac{1}{2} I_B \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega^2 = \frac{1}{4} M R^2 \omega^2 \\$• Work done for the circular ring (C):$\\ W_C = \frac{1}{2} I_C \omega^2 = \frac{1}{2} \left(M R^2\right) \omega^2 = \frac{1}{2} M R^2 \omega^2 \\$Now, compare the work done for each object:$\\ W_C > W_B > W_A \\$Therefore, the correct relation is:$\\$Option 3: $W_C > W_B > W_A \\$This corresponds to Option 3.