$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $L_0$ to $4L_0$ in $4 \ s$. The magnitude of this torque is

The moment of inertia of a wheel about its vertical axis is $2 \ kg \ m^2$. It is rotating about this axis at a speed of $60 \ rpm$. What torque must be applied to bring the wheel to rest in $1 \ minute$?

$A$ disc has mass $M$ and radius $R$. How much tangential force should be applied to the rim of the disc so as to rotate the disc with angular velocity $\omega$ in time $t$?

$A$ 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:

$A$ triangular plate of uniform thickness and density rotates about an axis perpendicular to the plane of the paper. In case $(a)$,the axis passes through point $A$,and in case $(b)$,it passes through point $B$. If an equal force $F$ is applied as shown in the figure,in which case will the angular acceleration be greater?

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be: