What is the value of the tangential component of the linear acceleration of a particle of a rigid body rotating with a constant angular velocity?

$A$ 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

$A$ particle starting from rest moves along the circumference of a circle of radius $r$ with angular acceleration $\alpha$. The magnitude of the average velocity in the time it completes the small angular displacement $\theta$ is

$A$ wheel,initially at rest,is rotated with a uniform angular acceleration. The wheel rotates through an angle ${\theta _1}$ in the first $1 \ s$ and through an additional angle ${\theta _2}$ in the next $1 \ s$. The ratio $\frac{{\theta _2}}{{\theta _1}}$ is

The ratio of the angular speed of a second-hand to the hour-hand of a watch is (in $: 1$)

$A$ wheel undergoes a constant angular acceleration from time $t=0$ to $t=20 \ s$ and thereafter angular acceleration is zero. If angular velocity at $t=2 \ s$ is found to be $5 \ rad/s$,then the number of revolutions made by the wheel in the time interval $t=0 \ s$ to $t=50 \ s$ is