$A$ wheel has a moment of inertia $5 \times 10^{-3} \, kg \, m^2$ and is making $20 \, rev/sec$. The torque needed to stop it in $10 \, sec$ is $............. \times 10^{-2} \, N-m$. (in $\pi$)

$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $L_0$ to $4L_0$ in $4 \ s$. The magnitude of this torque is

$A$ 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

In the given figure,$r_1$ and $r_2$ are $10 \ cm$ and $20 \ cm$ respectively. If the moment of inertia of the wheel is $1500 \ kg \cdot m^2$,find the angular acceleration.

$A$ 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 $rad/s$?

$A$ solid disc of radius $20 \, \text{cm}$ and mass $10 \, \text{kg}$ is rotating with an angular velocity of $600 \, \text{rpm}$ about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc to rest in $10 \, \text{s}$ is $x \pi \times 10^{-1} \, \text{Nm}$. Find the value of $x$.