Here, radius of both the ring and solid disc $R = 10 \,\text{cm} = 0.1 \,\text{m}, \mu_k = 0.2$Moment of inertia of the solid disc $= \frac{1}{2} M R^2$Initial angular velocity $\omega_0 = 10 \pi \,\text{rad}\,\text{s}^{-1}$Initial velocity of centre of mass is zero. Frictional force causes the C.M. to accelerate $F = m a ; m a = \mu_{k} m g$ $a = \mu_{k} g \ldots (i)$Now, $v = u + a t$ $v = \mu_{k} g t \ (\because u = 0) \ldots (ii)$Torque due to friction causes retardation in the initial angular speed $\omega_0$ $\mu_{k} m g \times R = -I \alpha$ $\alpha = \frac{-\mu_{k} m g R}{I} \ldots (iii)$ $\text{As } \omega = \omega_0 + \alpha t$ $\therefore \omega = \omega_0 - \frac{\mu_{k} m g R t}{I} \ldots (iv)$Rolling begins, when $v = R \omega$From (ii) and (iv), $\mu_{k} g t = R \omega_0 - \frac{\mu_{k} m g R^{2} t}{I} \ldots (v)$For a ring $I = m R^2$$\therefore \mu_{k} g t = R \omega_0 - \frac{\mu_{k} m g R^{2} t}{m R^{2}}$$\mu_{k} g t = R \omega_0 - \mu_{k} g t$; $2 \mu_{k} g t = R \omega_0$$t_1 = \frac{R \omega_0}{2 \mu_{k} g} \ldots (vi)$For a disc, $I = \frac{1}{2} m R^2$∴ From equation (v) $\mu_{k} g t = R \omega_0 - \frac{\mu_{k} m g R^{2} t}{\frac{1}{2} m R^{2}}$$\mu_{k} g t = R \omega_0 - 2 \mu_{k} g t$$3 \mu_{k} g t = R \omega_0 \ldots (vii)$$t_2 = \frac{R \omega_0}{3 \mu_{k} g}$Putting $R = 0.1 \,\text{m}, \omega_0 = 10 \pi \,\text{rad}\,\text{s}^{-1},$$\mu_k = 0.2$$g = 9.8 \,\text{m}\,\text{s}^{-2}$ in equation (vi) and (vii)$t_1 = \frac{0.1 \times 10 \pi}{2 \times 0.2 \times 9.8} = 0.8 \,\text{s}$$t_2 = \frac{0.1 \times 10 \pi}{3 \times 0.2 \times 9.8} = 0.53 \,\text{s}$It is clear that $t_2 < t_1$, so disc will start rolling first.