$A$ force of $1 \ N$ is applied at the free edge of a door of width $1.6 \ m$ to open it. If the door is to be opened by applying force at a point $0.4 \ m$ away from the hinges (axis of rotation),then the force required is ...... $N$.

$A$ disc of mass $10 \ kg$ and radius $0.1 \ m$ is rotating at $120 \ rpm$. $A$ retarding torque brings it to rest in $10 \ s$. If the same torque is due to a force applied tangentially on the rim of the disc, then the magnitude of the force is: (in $\pi \ N$)

$A$ string is wrapped around the rim of a wheel of moment of inertia $0.20\, kg-m^2$ and radius $20\, cm$. The wheel is free to rotate about its axis and initially the wheel is at rest. The string is now pulled by a force of $20\, N$. The angular velocity of the wheel after $5\, s$ will be ....... $rad/s$.

$A$ wheel of radius $0.4 \,m$ can rotate freely about its axis as shown in the figure. $A$ string is wrapped over its rim and a mass of $4 \,kg$ is hung. An angular acceleration of $8 \,rad \,s^{-2}$ is produced in it due to the torque. Then, the moment of inertia of the wheel is $(g = 10 \,m \,s^{-2})$.

$A$ uniform disc of mass $5\,g$ and radius $1\,cm$ is fixed to a thin stick $AB$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque,that will make the system rotate about $AB$ at $25$ rotations per second in $5\,s$,is close to:

$A$ 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)