The moment of inertia of a wheel about its vertical axis is $2 \ kg \ m^2$. It is rotating about this axis at a speed of $60 \ rpm$. What torque must be applied to bring the wheel to rest in $1 \ minute$?

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:

$A$ thin uniform hemispherical bowl of mass $m$ and radius $R$ is lying on a smooth horizontal surface. $A$ horizontal force $F$ is now applied perpendicular to the rim of the bowl (see figure). The instantaneous angular acceleration of the bowl will be

$A$ string is wrapped around the rim of a wheel of moment of inertia $0.40 \ kg \cdot m^2$ and radius $10 \ cm$. The wheel is free to rotate about its axis. Initially,the wheel is at rest. The string is now pulled by a force of $40 \ N$. The angular velocity of the wheel after $10 \ s$ is $x \ rad/s$,where $x$ is $\qquad$

$A$ uniform rod of mass $M$ and length $L$ is hinged at one end as shown in the figure. Initially,the rod is kept horizontal by a massless string. When the string is cut,the angular acceleration of the rod is:

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