$A$ uniform rod of length $l$, hinged at the lower end is free to rotate in the vertical plane . If the rod is held vertically in the beginning and then released, the angular acceleration of the rod when it makes an angle of $45^o$ with the horizontal ($I = ml^2/3$)

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 moment of inertia $I$ of a body rotating about an axis with angular speed $\omega $ increases, its angular momentum $L$ is unchanged but the kinetic energy $K$ increases if there is no torque applied on it.
Statement $2$ : $L = I\omega $, kinetic energy of rotation $ = \frac{1}{2}\,I\omega ^2$

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 \;kg – m ^{2}$ and $10\; rad \,s^{-1}$ respectively while those for the second one are $0.2 \;kg – m ^{2}$ and $5\; rad \,s ^{-1}$ respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is ………..$J$

The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is $2 : 1$. If their radii are in the ratio $2 : 1$; then the ratio of their masses will be

$A$ uniform rod of mass $M$ is hinged at its upper end. $A$ particle of mass $m$ moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision find the value of $M/m =?$