In the following figure,$r_1 = 5\,cm$ and $r_2 = 30\,cm$. If the moment of inertia of the wheel is $1500\,kg\cdot m^2$,then its angular acceleration will be (approximately):

$A$ flywheel of mass $25 \,kg$ has a radius of $0.2 \,m$. It is rotating at $240 \,rpm$. What is the torque necessary to bring it to rest in $20 \,s$?

$A$ uniform disc rotates at $10$ revolutions per second. $A$ torque is applied to it, producing an angular acceleration of $5 \text{ rad s}^{-2}$. After $2 \text{ s}$, its angular velocity is ...... $\text{rad s}^{-1}$ and the number of revolutions completed by the disc in $2 \text{ s}$ 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 ............

$A$ 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 is:

$A$ thin uniform rod of mass $M$ and length $L$ is pivoted at a height $\frac{L}{3}$ from its lower end as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is . . . . . . . ($g$ = gravitational acceleration)