The thin rod shown below has mass $M$ and length $L$. $A$ force $F$ acts at one end as shown,and the rod is free to rotate about the other end in the plane of the force. The initial angular acceleration of the rod is ........

The moment of inertia of a wheel about an axis passing through its center is $200 \, kg \cdot m^2$. $A$ constant torque of $1000 \, N \cdot m$ is applied to rotate the wheel. After $3 \, s$,the angular velocity of the wheel will be ........ $rad/s$.

$A$ 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 $\pi \,N$)

The moment of inertia of a body about a given axis is $1.2 \ kg \cdot m^2$. Initially,the body is at rest. To produce a rotational kinetic energy of $1500 \ J$,an angular acceleration of $25 \ rad/s^2$ must be applied to the body for a duration of ........ $(s)$.

The moment of inertia of a body about a given axis is $2.4 \, kg \cdot m^2$. For what duration of time must an angular acceleration of $5 \, rad/s^2$ be applied about this axis to generate a rotational kinetic energy of $750 \, J$? (in seconds)

Which of the following are correct expressions for torque acting on a body?
$A. \ \vec{\tau}=\vec{ r } \times \vec{ L }$
$B. \ \vec{\tau}=\frac{ d }{ dt }(\vec{ r } \times \vec{ p })$
$C. \ \vec{\tau}=\vec{ r } \times \frac{ d \vec{ p }}{ dt }$
$D. \ \vec{\tau}= I \vec{\alpha}$
$E. \ \vec{\tau}=\vec{ r } \times \vec{ F }$
($\vec{ r }=$ position vector; $\vec{ p }=$ linear momentum;
$\vec{ L }=$ angular momentum; $\vec{\alpha}=$ angular acceleration;
$I=$ moment of inertia; $\vec{ F }=$ force; $t =$ time)
Choose the correct answer from the options given below: