$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $A_0$ to $4 \,A_0$ in $4 \,s$. The magnitude of the torque is

$A$ constant torque acting on a wheel changes its angular momentum from $A_0$ to $4A_0$ in $4 \ s$. The magnitude of the torque is:

$A$ wheel having a moment of inertia of $4 \,kg \cdot m^2$ about its symmetrical axis rotates at a rate of $240 \,rpm$ about it. The torque required to stop the rotation of the wheel in one minute is ............ $N \cdot m$.

The moment of inertia of a wheel rotating about an axis is $3 \times 10^2 \ kg \ m^2$ and it has a constant angular speed of $4.6 \ rad \ s^{-1}$. If a retarding torque of $6.9 \times 10^2 \ N \ m$ is applied to the wheel,in how many seconds will the wheel come to rest?

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