If there is a change of angular momentum from $1\,J\cdot s$ to $5\,J\cdot s$ in $5\,s$,then the torque is:

$A$ rectangular solid rod of length $0.3\, m$ is held horizontally,with one of its sides on the edge of a platform of height $5\, m$. When released,it slips off the table in a very short time $\Delta t = 0.01\, s$,remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

$A$ 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

$A$ torque of $30 \, N \cdot m$ is applied for $10 \, s$ on a wheel of mass $5 \, kg$ and moment of inertia $2 \, kg \cdot m^2$. The angular displacement of the wheel in $10 \, s$ will be ....... radians.

$A$ wheel is subjected to a torque of $1000 \ N-m$ and rotates with a moment of inertia of $200 \ kg-m^2$ about an axis passing through its center. What will be the angular velocity of the wheel after $3 \ s$?

$A$ 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 $x$.