$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of [where $g$ is gravitational acceleration]:

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:

Two discs of moment of inertia $I_1 = 4 \ kg \ m^2$ and $I_2 = 2 \ kg \ m^2$ about their central axes and normal to their planes,rotating with angular speeds $10 \ rad/s$ and $4 \ rad/s$ respectively,are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is . . . . . . $J$.

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that:

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:

During a test of a jet engine,its compressor rotates according to the graph shown. The total number of revolutions completed by the compressor during the test is: