For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is $\frac{x}{5}$. The value of $x$ is ................

Write the formula for rotational kinetic energy.

The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$, from rest without sliding, is

A  solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously.  The ratio $K_t : (K_t + K_r)$ for the sphere is

A disc of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:

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 ($2$)