$A$ 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 in newton is

$A$ wheel of mass $20 \,kg$ and radius $30 \,cm$ is rotating at an angular speed of $80 \,rev/min$ when the motor is turned off. Neglecting the friction at the axis, calculate the force that must be applied tangentially to the wheel to bring it to rest in $5 \,rev$. (in $\pi \,N$)

$A$ 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

$A$ thin solid disk of $1 \ kg$ is rotating along its diameter axis at the speed of $1800 \ rpm$. By applying an external torque of $25 \pi \ Nm$ for $40 \ s$,the speed increases to $2100 \ rpm$. The diameter of the disk is . . . . . . $m$.

$A$ constant torque of $1000 \; Nm$ turns a wheel of moment of inertia $200 \; kg \cdot m^2$ about an axis through its centre. The wheel is at rest initially. Its angular velocity after $3 \; s$ is

$A$ rigid body rotates about a fixed axis with variable angular velocity $\omega(t) = \alpha - \beta t$ at time $t$,where $\alpha$ and $\beta$ are constants. The angle through which it rotates before it comes to rest is