$A$ particle of mass $m$ moves in a circular path of radius $r$,under the action of a force which delivers constant power $P$ and increases its speed. The angular acceleration of the particle at time $t$ is proportional to:

Two small balls with masses $m$ and $2m$ are attached to both ends of a rigid rod of length $d$ and negligible mass. If the angular momentum of this system is $L$ about an axis $(A)$ passing through its centre of mass and perpendicular to the rod,then the angular velocity of the system about $A$ is:

$A$ body rotating at $20 \ rad/s$ is acted upon by a constant torque providing it a deceleration of $2 \ rad/s^2$. At what time will the body have kinetic energy same as the initial value if the torque continues to act?

Obtain the relation between torque and moment of inertia.

The moment of inertia of a body about a given axis is $1.2 \; kg \cdot m^2$. Initially,the body is at rest. In order to produce a rotational kinetic energy of $1500 \; J$,an angular acceleration of $25 \; rad/s^2$ must be applied about that axis for a duration of: (in $; s$)

$A$ string is wrapped around the rim of a wheel of moment of inertia $0.40 \ kg \cdot m^2$ and radius $10 \ cm$. The wheel is free to rotate about its axis. Initially,the wheel is at rest. The string is now pulled by a force of $40 \ N$. The angular velocity of the wheel after $10 \ s$ is $x \ rad/s$,where $x$ is $\qquad$