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:

$A$ uniform disc is rolling on a horizontal surface. At a certain instant,$B$ is the point of contact and $A$ is the topmost point of the disc,where $R$ is the radius of the disc.

The graphs below show angular velocity $\omega$ as a function of time $t$. In which one is the magnitude of the angular acceleration $\alpha$ constantly decreasing?

$A$ solid sphere of mass $m$,radius $R$,having moment of inertia about an axis passing through its center of mass as $I$ is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through the rim (edge) and perpendicular to its plane remains $I$. Then the radius of the disc is:

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

One of two identical cylinders,cylinder $A$,rotates at an angular speed of $50 \text{ revolutions per second}$. This rotating cylinder is brought into contact with a second stationary cylinder,$B$. Due to kinetic friction between the two cylinders,the stationary cylinder starts rotating with an angular acceleration,while cylinder $A$ undergoes angular deceleration. If the magnitude of the angular acceleration for both cylinders is $1 \text{ revolution per second}^2$,after how many seconds $(t)$ will the angular speeds of both cylinders become equal?