If $I = 50\,kg-m^2$,then how much torque (in $N-m$) must be applied to stop it in $10\,s$,given its initial angular speed is $20\,rad/s$?

$A$ ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off,the angular velocity becomes $\left(\frac{1}{4}\right)^{th}$ of the original in time $t$ and $n$ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest is (Angular retardation is uniform):

$A$ 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 length $l$ and mass $m$ is pivoted at point $A$. The rod is released from a horizontal position. If the moment of inertia of the rod about point $A$ is $ml^2/3$,then its initial angular acceleration is:

$A$ rectangular plate of mass $20 \ kg$ is suspended from points $A$ and $B$ as shown in the figure. If the pin at $B$ is suddenly removed,find the angular acceleration of the plate in $rad/s^2$.

$A$ uniform metre stick of mass $M$ and length $L = 1 \ m$ is hinged at one end and supported in a horizontal direction by a string attached to the other end. What should be the initial angular acceleration of the free end of the stick if the string is cut? (in $rad/s^2$)