A disc and a ring of same mass are rolling and if their kinetic energies are equal, then the ratio of their velocities will be

The total kinetic energy of a body of mass $10\ kg$ and radius $0.5\ m$ moving with a velocity of $2\ m/s$ without slipping is $32.8\ joule$.  The radius of gyration of the body is  ………. $m$

Rotational kinetic energy of a given body about an axis is proportional to

A $70\, kg$ man leaps vertically into the air from a crouching position. To take the leap the man pushes the ground with a constant force $F$ to raise himself The center of gravity rises by $0.5\, m$ before he leaps. After the leap the $c.g.$ rises by another $1\, m$. The maximum power delivered by the muscles is : (Take $g\, = 10\, ms^{-2}$)

One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure $1$ . In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $\mathrm{r}$. The finger rotates with an angular velocity $\omega_0$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure $2$). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.

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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 ($2$)