$A$ wheel having a moment of inertia $2 \; kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \; rpm$ about this axis. The torque required to stop the wheel's rotation in one minute is:

The moment of inertia of a wheel about its transverse axis is $2 \ kg \cdot m^2$ and it is rotating at $60 \ rpm$ about that axis. The torque required to stop the wheel in one minute is:

$A$ stationary horizontal disc is free to rotate about its axis. When a torque is applied on it,its kinetic energy as a function of $\theta,$ where $\theta$ is the angle by which it has rotated,is given as $K\theta^2.$ If its moment of inertia is $I,$ then the angular acceleration of the disc is

$A$ wheel is rotating about an axis through its centre at $720$ revolutions per minute. It is acted on by a constant torque opposing its motion for $8$ seconds until it comes to rest. The number of revolutions made by the wheel in the $8$ seconds is:

$A$ disc of radius $0.4$ m and mass $1$ kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration is $10$ rad/s$^2$. The tangential force applied to the rim of the disc is (in $N$)

$A$ slender uniform rod of mass $M$ and length $l$ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The rod is released from a position where it makes an angle $\theta$ with the vertical. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is