$A$ uniform thin rod of length $2L$ and mass $m$ lies on a horizontal table. $A$ horizontal impulse $J$ is given to the rod at one end. There is no friction. The total kinetic energy of the rod just after the impulse will be

Two equal masses each of mass $M$ are joined by a massless rod of length $L$. Now an impulse $J = MV$ is given to one of the masses,making an angle of $30^{\circ}$ with the length of the rod. The angular velocity of the rod just after imparting the impulse is:

In the case of rotational dynamics,which one of the following statements is correct?
$[\vec{\omega} = \text{angular velocity}, \vec{v} = \text{linear velocity}, \vec{r} = \text{radius vector}, \vec{\alpha} = \text{angular acceleration}, \vec{a} = \text{linear acceleration}, \vec{L} = \text{angular momentum}, \vec{p} = \text{linear momentum}, \vec{\tau} = \text{torque}, \vec{f} = \text{force}]$

The angular momentum of a rotating body is $L$. When the frequency of the rotating body is tripled and its kinetic energy is made one-third,the new angular momentum becomes:

One ice skater of mass $m$ moves with speed $2v$ to the right,while another of the same mass $m$ moves with speed $v$ toward the left,as shown in figure $I$. Their paths are separated by a distance $b$. At $t = 0$,when they are both at $x = 0$,they grasp a pole of length $b$ and negligible mass. For $t > 0$,consider the system as a rigid body of two masses $m$ separated by distance $b$,as shown in figure $II$. Which of the following is the correct formula for the motion after $t = 0$ of the skater initially at $y = b/2$?

The moment of inertia of a sphere of mass $M$ and radius $R$ is $I$. If the mass is kept constant,what will be the nature of the graph between $I$ and $R$?