The instantaneous angular position of a rotating wheel is given by the formula $\theta (t) = 2t^3 - 6t^2$. At what time will the torque on this wheel be zero? $t = \dots \text{s}$

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be:

$A$ disc of mass $25 \ kg$ and radius $0.2 \ m$ is rotating at $240 \ r.p.m.$ $A$ retarding torque brings it to rest in $20 \ s$. If the torque is due to a force applied tangentially on the rim of the disc,then the magnitude of the force is:

$A$ uniform thin rod of length $L$ and mass $m$ is lying on a smooth horizontal table. $A$ horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

$A$ billiard ball is hit by a cue at a point distance $h$ above the centre. It acquires a linear velocity $v_0$. Let $m$ be the mass and $r$ be the radius of the ball. The angular velocity acquired by the ball is

At time $t = 0$,a $2\, kg$ particle has position vector $\vec{r} = (4\hat{i} - 2\hat{j})\, m$ relative to the origin. Its velocity is given by $\vec{v} = 2t^2\hat{i}\, m/s$. The torque acting on the particle about the origin at $t = 2\, s$ is ........ $\hat{k}\, N\cdot m$.