$A$,$B$,and $C$ are a disc,a solid sphere,and a spherical shell respectively,with the same radii $(R)$ and masses $(M)$. These bodies are placed as shown in the figure. The moment of inertia of the given system about the axis $PQ$ is $\frac{x}{15} I$,where $I$ is the moment of inertia of the disc about its diameter. The value of $x$ is . . . . . . .

If the earth suddenly contracts to $(1/3)^{rd}$ of its present size without any change in its mass, the ratio of the kinetic energy of the earth after and before contraction will be (Earth is assumed to be a rotating sphere about itself).

$A$ rotating body has angular momentum $L$. If its frequency is doubled and its kinetic energy is halved,what will be its new angular momentum?

Fill in the blanks:
$(1)$ If $|\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B}$,then the angle between $\vec{A}$ and $\vec{B}$ is ............ .
$(2)$ The angle between the angular momentum and linear momentum of a particle in rotational motion is ............ .
$(3)$ $A$ force $F\hat{k}$ acts on a particle having position vector $(2\hat{i} + \hat{j})$. The torque acting on the particle is ............ .

$A$ uniform circular disc of radius $R$, lying on a frictionless horizontal plane, is rotating with an angular velocity $\omega$ about its own axis. Another identical circular disc is gently placed on top of the first disc coaxially. The loss in rotational kinetic energy due to friction between the two discs, as they acquire a common angular velocity, is ($I$ is the moment of inertia of the disc).

$A$ rod of mass $M$ and length $l$ is at rest on a smooth horizontal surface. $A$ particle of the same mass $M$ strikes one end of the rod with velocity $u$ perpendicular to the rod,elastically. Just after the collision,what is the kinetic energy of the upper half part of the rod?