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(D) $1$. When the ring enters the rough surface,friction acts backwards,causing linear deceleration: $a = \frac{f}{M} = \frac{\mu Mg}{M} = \mu g$. The

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All of the above.

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(D) $1$. When the ring enters the rough surface,friction acts backwards,causing linear deceleration: $a = \frac{f}{M} = \frac{\mu Mg}{M} = \mu g$. The torque about the centre of mass is $	au = fR = \mu MgR$. Angular acceleration $\alpha = \frac{	au}{I} = \frac{\mu MgR}{MR^2} = \frac{\mu g}{R}$.$2$. Rolling starts when $v = \omega R$. At time $t$,$v = v_0 - \mu gt$ and $\omega = \alpha t = \frac{\mu g}{R}t$. Setting $v_0 - \mu gt = (\frac{\mu g}{R}t)R = \mu gt$,we get $t = \frac{v_0}{2\mu g}$.$3$. Distance moved $s = v_0 t - \frac{1}{2}at^2 = v_0(\frac{v_0}{2\mu g}) - \frac{1}{2}(\mu g)(\frac{v_0}{2\mu g})^2 = \frac{v_0^2}{2\mu g} - \frac{v_0^2}{8\mu g} = \frac{3v_0^2}{8\mu g}$. Option $(A)$ is correct.$4$. Final velocity $v = v_0 - \mu g(\frac{v_0}{2\mu g}) = \frac{v_0}{2}$. Final angular velocity $\omega = \frac{v}{R} = \frac{v_0}{2R}$.$5$. Gain in rotational kinetic energy $\Delta K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(MR^2)(\frac{v_0}{2R})^2 = \frac{Mv_0^2}{8}$. Option $(B)$ is correct.$6$. Initial kinetic energy $K_i = \frac{1}{2}Mv_0^2$. Final kinetic energy $K_f = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}M(\frac{v_0}{2})^2 + \frac{1}{2}(MR^2)(\frac{v_0}{2R})^2 = \frac{Mv_0^2}{8} + \frac{Mv_0^2}{8} = \frac{Mv_0^2}{4}$.$7$. Loss in kinetic energy $\Delta K = K_i - K_f = \frac{1}{2}Mv_0^2 - \frac{1}{4}Mv_0^2 = \frac{1}{4}Mv_0^2$. Option $(C)$ is correct.

$A$ ring of mass $M$ and radius $R$ sliding with a velocity $v_0$ suddenly enters a rough surface where the coefficient of friction is $\mu$,as shown in the figure. Choose the correct alternative$(s)$.

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