The linear speed of a particle at the equator of the earth due to its spin motion is ' V '. The linear speed of the particle at latitude $30^{\circ}$ is

$$\left[\begin{array}{l} \sin 30^{\circ}=\cos 60^{\circ}=1 / 2 \\ \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2 \end{array}\right]$$

Two objects of masses ' $m_1$ ' and ' $m_2$ ' are moving in the circles of radii ' $r_1$ ' and ' $r_2$ ' respectively. Their respective angular speeds ' $\omega_1$ ' and ' $\omega_2$ ' are such that they both complete one revolution in the same time ' $t$ '. The ratio of linear speed of ' $m_2$ ' to that of ' $m_1$ ' is

A body performing uniform circular motion of radius ' $R$ ' has frequency ' $n$ '. Its centripetal acceleration per unit radius is proportional to $(n)^x$. The value of $x$ is

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 time it completes the small angular displacement ' $\theta$ ' is