Moment of inertia of a body about a given axis is $1.5\, kg\, m^2$. Initially,the body is at rest. In order to produce a rotational kinetic energy of $1200\, J$,an angular acceleration of $20\, rad/s^2$ must be applied about the axis of rotation for a duration of ......... $\sec$.

$A$ wheel with a moment of inertia of $5 \times 10^{-3} \ kg \cdot m^2$ is rotating at a rate of $20 \ rev/s$. The angular deceleration required to bring it to rest in $20 \ s$ is:

$A$ wheel of radius $0.2 \ m$ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $10 \ N$ as shown in the figure. The established torque produces an angular acceleration of $2 \ rad/s^2$. The moment of inertia of the wheel is . . . . . . $kg \ m^2$. (Acceleration due to gravity $= 10 \ m/s^2$)

$A$ wheel is rotating about an axis through its centre at $720$ revolutions per minute. It is acted on by a constant torque opposing its motion for $8$ seconds until it comes to rest. The number of revolutions made by the wheel in the $8$ seconds is:

$A$ uniform disc of mass $m$ and radius $R$ is pivoted at point $P$ and is free to rotate in a vertical plane. The centre $C$ of the disc is initially in a horizontal position with $P$ as shown in the figure. If it is released from this position,then its angular acceleration when the line $PC$ is inclined to the horizontal at an angle $\theta$ is ............

$A$ string is wrapped around the rim of a wheel with a moment of inertia of $0.20 \ kg \cdot m^2$ and a radius of $20 \ cm$. The wheel is free to rotate about its axis and is initially at rest. The string is now pulled with a force of $20 \ N$. What will be the angular velocity of the wheel after $5 \ s$ in $rad/s$?