$A$ torque of $50\, Nm$ acting on a wheel at rest rotates it through $200\, rad$ in $5\, s$. Calculate the angular acceleration produced. ........... $rad\, s^{-2}$.

$A$ thin solid disk of $1 \ kg$ is rotating along its diameter axis at the speed of $1800 \ rpm$. By applying an external torque of $25 \pi \ Nm$ for $40 \ s$,the speed increases to $2100 \ rpm$. The diameter of the disk is . . . . . . $m$.

$A$ disc of mass $25 \ kg$ and radius $0.2 \ m$ is rotating at $240 \ r.p.m.$ $A$ retarding torque brings it to rest in $20 \ s$. If the torque is due to a force applied tangentially on the rim of the disc,then the magnitude of the force in newton is

$A$ 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:

The moment of inertia of a body about a given axis is $2.4 \, kg \cdot m^2$. For what duration of time must an angular acceleration of $5 \, rad/s^2$ be applied about this axis to generate a rotational kinetic energy of $750 \, J$? (in seconds)

$A$ body rotates about a fixed axis with an angular acceleration of $3 \, rad/s^2$. The angle rotated by it during the time when its angular velocity increases from $10 \, rad/s$ to $20 \, rad/s$ (in radian) is ..........