This question has Statement $1$ and Statement $2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1$ : When the moment of inertia $I$ of a body rotating about an axis with angular speed $\omega$ increases,its angular momentum $L$ remains unchanged,but the kinetic energy $K$ decreases if no external torque is applied.
Statement $2$ : $L = I\omega$ and the rotational kinetic energy $K = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$.

The moment of inertia of a rectangular plate about an axis passing through $P$ and perpendicular to the plate is $I$. Then,the moment of inertia of the triangular portion $PQR$ about an axis perpendicular to the plane of the plate and passing through $P$ is:

$A$ body kept on a smooth horizontal surface is pulled by a constant horizontal force applied at the top point of the body. If the body rolls purely on the surface,its shape can be:

The figure shows a system consisting of $(i)$ a ring of outer radius $3R$ rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and $(ii)$ an inner disc of radius $2R$ rotating anti-clockwise with angular speed $\omega/2$. The ring and disc are separated by frictionless ball bearings. The system is in the $x-z$ plane. The point $P$ on the inner disc is at distance $R$ from the origin,where $OP$ makes an angle of $30^{\circ}$ with the horizontal. Then with respect to the horizontal surface,
$(A)$ the point $O$ has linear velocity $3R\omega\hat{i}$.
$(B)$ the point $P$ has a linear velocity $\frac{11}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}$.
$(C)$ the point $P$ has linear velocity $\frac{13}{4}R\omega\hat{i} - \frac{\sqrt{3}}{4}R\omega\hat{k}$.
$(D)$ The point $P$ has a linear velocity $(3 - \frac{\sqrt{3}}{4})R\omega\hat{i} + \frac{1}{4}R\omega\hat{k}$.

$A$ uniform rod of length $L$ and mass $M$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ moving with velocity $v$ strikes the rod at one end perpendicular to the rod. After the collision,the particle comes to rest. What is the angular velocity of the rod about its centre of mass after the collision?

The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of the particle is doubled and the frequency is halved,then the new angular momentum becomes: