$A$ particle is revolving in a circle of radius $2 \ m$ with angular velocity $\omega = t^2 - 4t + 8 \ rad/s$. The time when the speed of the particle becomes $8 \ m/s$ is ......... $\sec$.

$A$ clock has a continuously moving second's hand of $0.1\, m$ length. The average acceleration of the tip of the hand (in units of $m\, s^{-2}$) is of the order of

$A$ body is revolving with a uniform speed $v$ in a circle of radius $r$. The tangential acceleration is

If a wheel starting from rest is rotating with an angular acceleration of $\pi \ rad \ s^{-2}$,then the number of rotations made by the wheel in the first $6 \ s$ is:

$A$ ball is spun with angular acceleration $\alpha = 6t^2 - 2t$,where $t$ is in seconds and $\alpha$ is in $rad/s^2$. At $t = 0$,the ball has an angular velocity of $10 \ rad/s$ and an angular position of $4 \ rad$. The most appropriate expression for the angular position of the ball is:

$A$ particle starting from rest moves in a circle of radius $r$. It attains a velocity of $V_{0} \; m/s$ in the $n^{\text{th}}$ round. Its angular acceleration will be