$A$ solid disc and a ring,both of radius $10\; cm$,are placed on a horizontal table simultaneously,with an initial angular speed equal to $10\pi\; rad\; s^{-1}$. Which of the two will start to roll earlier? The coefficient of kinetic friction is $\mu_k = 0.2$.

(B) Given: Radius $r = 10\; cm = 0.1\; m$,initial angular speed $\omega_0 = 10\pi\; rad\; s^{-1}$,coefficient of kinetic friction $\mu_k = 0.2$,initial linear velocity $u = 0$.
The frictional force $f = \mu_k mg$ provides linear acceleration $a = \mu_k g$. Using $v = u + at$,we get $v = \mu_k gt$.
The torque $\tau = -f r = -I\alpha$ causes angular deceleration $\alpha = \frac{\mu_k mgr}{I}$. Using $\omega = \omega_0 - \alpha t$,we get $\omega = \omega_0 - \frac{\mu_k mgr}{I}t$.
Rolling starts when $v = r\omega$. Substituting the expressions:
$\mu_k gt = r(\omega_0 - \frac{\mu_k mgr}{I}t) \implies \mu_k gt = r\omega_0 - \frac{\mu_k mgr^2}{I}t$.
For the ring $(I = mr^2)$:
$\mu_k gt_r = r\omega_0 - \mu_k gt_r \implies 2\mu_k gt_r = r\omega_0 \implies t_r = \frac{r\omega_0}{2\mu_k g} = \frac{0.1 \times 10\pi}{2 \times 0.2 \times 9.8} \approx 0.80\; s$.
For the disc $(I = \frac{1}{2}mr^2)$:
$\mu_k gt_d = r\omega_0 - 2\mu_k gt_d \implies 3\mu_k gt_d = r\omega_0 \implies t_d = \frac{r\omega_0}{3\mu_k g} = \frac{0.1 \times 10\pi}{3 \times 0.2 \times 9.8} \approx 0.53\; s$.
Since $t_d < t_r$,the disc starts rolling earlier than the ring.