$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

$A$ uniform rod of mass $m$ and length $2l$ is balanced on a triangular prism. Now,a length of $l/2$ is cut from one end of the rod and placed over the shortened part such that the ends meet. The initial angular acceleration is:

The angular velocity of a body changes from $6 \ rad \ s^{-1}$ to $21 \ rad \ s^{-1}$ in a time of $1.5 \ s$. If the moment of inertia of the body is $100 \ g \ m^2$,then the rate of change of angular momentum of the body is (in $N \ m$)

$A$ uniform disc of mass $5\,g$ and radius $1\,cm$ is fixed to a thin stick $AB$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque,that will make the system rotate about $AB$ at $25$ rotations per second in $5\,s$,is close to:

$A$ rod of length $50\,cm$ is pivoted at one end. It is raised such that it makes an angle of $30^o$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$) will be $(g = 10\,ms^{-2})$.

An air compressor is powered by a $200\,rad\,s^{-1}$ electric motor using a $V-$belt drive. The motor pulley is $8\,cm$ in radius,and the tension in the $V-$belt is $135\,N$ on one side and $45\,N$ on the other. The power of the motor will be ...... $kW$.