- A$0.4$
- ✓$0.2$
- C$0.6$
- D$0.8$
$\therefore \mu = \frac{{{u^2}}}{{2gs}} = \frac{{{{(6)}^2}}}{{2 \times 10 \times 9}} = 0.2$
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($1$) The total kinetic energy of the ring is
$[A]$ $\mathrm{M} \omega_0^2 \mathrm{R}^2$ $[B]$ $\frac{1}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$ $[C]$ $\mathrm{M \omega}_0^2(\mathrm{R}-\mathrm{r})^2$ $[D]$ $\frac{3}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$
($2$) The minimum value of $\omega_0$ below which the ring will drop down is
$[A]$ $\sqrt{\frac{g}{\mu(R-r)}}$ $[B]$ $\sqrt{\frac{2 g}{\mu(R-r)}}$ $[C]$ $\sqrt{\frac{3 g}{2 \mu(R-r)}}$ $[D]$ $\sqrt{\frac{g}{2 \mu(R-r)}}$
Givin the answer quetion ($1$) and ($2$)

