When $12000 \ J$ of work is done on a flywheel,its frequency of rotation increases from $10 \ Hz$ to $20 \ Hz$. The moment of inertia of the flywheel about its axis of rotation is (take $\pi^2=10$):

$A$ cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of the wheel is $10 \ kg$ and the radius is $10 \ cm$. It can rotate freely without any friction. Initially,the wheel is at rest. If a steady pull of $20 \ N$ is applied on the cord,the angular velocity (in $rad/s$) of the wheel,after the cord is unwound by $1 \ m$,would be:

$A$ wheel is rotating with an angular speed of $20\ rad/s$. It is stopped to rest by applying a constant torque in $4\ s$. If the moment of inertia of the wheel about its axis is $0.20\ kg\cdot m^2$,then the work done by the torque in two seconds will be .......... $J$.

$A$ uniform bar of length $6l$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speeds $2v$ and $v$ respectively,strike the bar (as shown in the figure) and stick to the bar after the collision. The total energy after the collision (about the center of mass,$c$) will be:

The $M.I.$ of a body about the given axis is $1.2 \, kg \cdot m^2$ and 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 ........ $s$.

The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?