According to the parallel axis theorem,$I = I_{cm} + Mx^2$. What will be the graph between $I$ and $x$?

$A$ body with a moment of inertia of $3 \ kg \cdot m^2$ rotating with an angular speed of $2 \ rad/s$ has the same kinetic energy as a mass of $12 \ kg$ moving with a speed of ......... $m/s$.

$A$ thin ring of mass $2 \ kg$ and radius $1 \ m$ is rolling without slipping on a horizontal plane with velocity $1 \ m/s$. $A$ small ball of mass $1 \ kg$,moving with velocity $2 \ m/s$ in the opposite direction,hits the ring at a height of $1.8 \ m$ and goes vertically up with velocity $1 \ m/s$. Immediately after the collision:
$(A)$ the ring has pure rotation about its stationary $CM$.
$(B)$ the ring comes to a complete stop.
$(C)$ friction between the ring and the ground is to the left.
$(D)$ there is no friction between the ring and the ground.

$A$ solid sphere of mass $M$ and radius $R$ has a moment of inertia $I$ about its diameter. It is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. The radius of the disc will be:

$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the ball starts moving up the inclined face of the wedge and rises to a height $h$. When the particle has risen to a height $h$ on the wedge,choose the correct alternative$(s)$.

$A$ spherical rigid ball is released from rest and starts rolling down an inclined plane from height $h=7 \, m$,as shown in the figure. It hits a block at rest on the horizontal plane (assume elastic collision). If the mass of both the ball and the block is $m$ and the ball is rolling without sliding,then the speed of the block after collision is close to ............. $m/s$.