$A$ uniform rod of mass $M$ and length $L$ is hinged at one end as shown in the figure. Initially,the rod is kept horizontal by a massless string. When the string is cut,the angular acceleration of the rod is:

$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $A_0$ to $4 \,A_0$ in $4 \,s$. The magnitude of the torque 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

The moment of inertia of a body about a given axis is $1.2 \ kg \cdot m^2$. Initially,the body is at rest. To produce a rotational kinetic energy of $1500 \ J$,an angular acceleration of $25 \ rad/s^2$ must be applied to the body for a duration of ........ $(s)$.

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be:

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane. One of its ends is hinged on a vertical axis. $A$ horizontal force $F = Mg/2$ is applied at a distance of $5L/6$ from the hinged end. What will be the angular acceleration of the rod?