$A$ wheel has a diameter of $1 \ m$. If it makes $30$ revolutions per second,then the linear speed of a point on its circumference will be:

The wheel of a car is rotating at the rate of $1200$ revolutions per minute. On pressing the accelerator for $10 \ s$,it starts rotating at $4500$ revolutions per minute. The angular acceleration of the wheel is:

$A$ wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first $2 \text{ s}$ it rotates through an angle $\theta_1$ and in the next $2 \text{ s}$ it rotates through an angle $\theta_2$. The ratio $\frac{\theta_2}{\theta_1}$ is . . . . . . .

$A$ particle moves along a circular path of radius $r = \frac{20}{\pi} \ m$ with a constant tangential acceleration. If the velocity of the particle at the end of the second revolution is $80 \ m/s$,what is the tangential acceleration?

$A$ disc starts from rest and acquires an angular velocity of $240 \ rpm$ in $10 \ s$. The angular acceleration is ...... $rad/s^2$ (assuming it is constant).

$A$ particle at rest starts moving with constant angular acceleration '$\alpha$' in a circular path of radius '$r$'. At a certain instant, the magnitude of centripetal acceleration is $(1/3)^{rd}$ the tangential acceleration. The relation between linear speed $(V)$ and angular acceleration $(\alpha)$ is