$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $A_0$ to $4 A_0$ in $4$ seconds. The magnitude of this torque is ...........

$A$ rod of length $50\,cm$ is pivoted at one end. It is raised such that it makes an angle of $30^o$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$) will be $(g = 10\,ms^{-2})$.

$A$ rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$ as shown in the figure. When the string is cut,the initial angular acceleration of the rod is

$A$ disc of radius $0.4 \,m$ and mass $1 \,kg$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration is $10 \,rad \,s^{-2}$. The tangential force applied to the rim of the disc is (in $\,N$)

$A$ wheel is subjected to a torque of $1000 \ N-m$ and rotates with a moment of inertia of $200 \ kg-m^2$ about an axis passing through its center. What will be the angular velocity of the wheel after $3 \ s$?

Write Newton's second law for a system of particles performing rotational motion.