$A$ torque of $10^5 \ N \cdot m$ is applied to a disc of mass $10 \ kg$ and radius of gyration $50 \ m$. What is the value of the angular acceleration in $rad/s^2$?

The angular speed of a motor wheel is increased from $1200 rpm$ to $3120 rpm$ in $16 s$. The angular acceleration of the motor wheel is

$A$ massless rod of length $2l$ has equal point masses $m$ attached to its two ends as shown in the figure. The rod is rotating about an axis passing through its centre and making an angle $\alpha$ with the axis. The magnitude of the rate of change of angular momentum of the rod,i.e.,$\left|\frac{dL}{dt}\right|$,is equal to:

$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

$A$ torque of $30 \ Nm$ is applied for $15 \ s$ on a stationary wheel of mass $5 \ kg$. The moment of inertia of the wheel is $2 \ kg \ m^2$. The angular displacement of the wheel in $10 \ s$ is ....... $rad$.

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be: