$A$ rope is wound around a solid cylinder of mass $1 \,kg$ and radius $0.4 \,m$. What is the angular acceleration of the cylinder, if the rope is pulled with a force of $25 \,N$? (The cylinder is rotating about its own axis.)

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:

$A$ 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 $3 \ s$?

Moment of inertia of a body about an axis is $4 \,kg-m^2$. The body is initially at rest and a torque of $8 \,N-m$ starts acting on it along the same axis. Work done by the torque in $20 \,s$, in joules, is

In the given figure,about which axis will the angular acceleration be greater if a force is applied at the midpoint of the triangular frame?

$A$ uniform disc of mass $m$ and radius $R$ is pivoted at point $P$ and is free to rotate in a vertical plane. The centre $C$ of the disc is initially in a horizontal position with $P$ as shown in the figure. If it is released from this position,then its angular acceleration when the line $PC$ is inclined to the horizontal at an angle $\theta$ is ............