$A$ flywheel gains a speed of $540\ r.p.m.$ in $6\ s$. Its angular acceleration will be

$A$ constant torque of $31.4 \, Nm$ is exerted on a pivoted wheel. If the angular acceleration of the wheel is $4\pi \, rad/s^2$,then the moment of inertia will be ....... $kg \cdot m^2$.

If the external torque acting on a system is $ \tau = 0 $, then:

$A$ 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

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be: