$A$ flywheel attains a speed of $540 \, r.p.m.$ in $6 \, s$. Its angular acceleration 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:

$A$ circular disc rotating with frequency $f_0 = 1.3 \, rev/sec$ comes to a stop in $30 \, seconds$. The approximate angular acceleration is ....... $rad/sec^2$.

If a constant torque acts on an object,then the object:

Find the ratio of linear acceleration $(a)$ to angular acceleration $(\alpha)$ of the center of mass $(COM)$ for the given diagram. Given: $m = 2 \, kg$,$r = 10 \, cm = 0.1 \, m$,and force $F = 20 \, N$ applied at the center.

$A$ 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 $8 \ s$?