$A$ thin uniform rod of mass $1 \ kg$ and length $1 \ m$ is hinged at one end to the ground. It originally stands vertically and is allowed to fall to the ground. If the rod hits the ground with angular speed $\omega$, then the correct statement is (Assume $g = 10 \ m \ s^{-2}$):

$A$ thin rod of length $l$ and mass $m$ oscillates in a vertical plane about a horizontal axis passing through one of its ends. If the maximum angular velocity of the rod is $\omega$,what is the maximum height reached by its center of mass?

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?

$A$ uniform rod of mass $M$ and length $L$ is pivoted at one end and is free to rotate in a vertical plane. The rod is released from rest in a horizontal position. What is the angular velocity of the rod when it reaches the vertical position?

$A$ disc of mass $1\,kg$ and radius $R$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of the disc. $A$ body of the same mass as that of the disc is fixed at the highest point of the disc. Now the system is released. When the body comes to the lowest position,its angular speed will be $4 \sqrt{\frac{x}{3 R}} \text{ rad s}^{-1}$ where $x=$ (Given $g = 10 \text{ m s}^{-2}$)

Explain the work done by torque.