$A$ flywheel of moment of inertia $2 \, kg \cdot m^2$ is rotated at a speed of $30 \, rad/s$. $A$ tangential force at the rim stops the wheel in $15 \, s$. The average torque of the force is ........... $N \cdot m$.

$A$ pulley has the shape of a uniform solid disc of mass $2 \ kg$ and radius $0.5 \ m$. $A$ string is wrapped over its rim and is pulled by a force of $2.5 \ N$. The pulley is free to rotate about its axis. Initially,the pulley is at rest. Find the angular velocity of the pulley just after $10 \ s$. (in $rad/s$)

$A$ wheel is at rest in a horizontal position. Its moment of inertia about the vertical axis passing through its centre is $I$. $A$ constant torque $\tau$ acts on it for $t$ seconds. The change in rotational kinetic energy is:

$A$ solid cylinder of mass $2\; kg$ and radius $4\; cm$ is rotating about its axis at the rate of $3\; rpm$. The torque required to stop after $2\pi$ revolutions is

$A$ 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:

If there is a change of angular momentum from $1\,J\cdot s$ to $5\,J\cdot s$ in $5\,s$,then the torque is: