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 ........

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:

$A$ uniform rod of mass $m$ and length $2l$ is balanced on a triangular prism. Now,a length of $l/2$ is cut from one end of the rod and placed over the shortened part such that the ends meet. The initial angular acceleration is:

$A$ disc of mass $25 \ kg$ and radius $0.2 \ m$ is rotating at $240 \ r.p.m.$ $A$ retarding torque brings it to rest in $20 \ s$. If the torque is due to a force applied tangentially on the rim of the disc,then the magnitude of the force in newton is

$A$ thin uniform rod of mass $M$ and length $L$ is pivoted at a height $\frac{L}{3}$ from its lower end as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is . . . . . . . ($g$ = gravitational acceleration)

$A$ circular disc of diameter $0.8 \ m$ and mass $4 \ kg$ is rolling on a smooth horizontal plane. If $2.56 \ N \ m$ torque is acting on the disc,then its angular acceleration is