$A$ particle moves in a circle of radius $25\, cm$ at two revolutions per second. The acceleration of the particle in $m/s^2$ is

$A$ point $P$ is moving in uniform circular motion with radius $3 \ m$. Let at some instant the acceleration of the point be $a = (6 \hat{i} - 4 \hat{j}) \ m/s^2$, the position vector be $r$, and the velocity vector be $v$. Choose the correct statement.

If the angular velocity of a merry-go-round is $60^o/s$ and you are $3.5\,m$ from the centre of rotation,your linear velocity will be

$A$ particle moves along a circular path of radius $r$ with uniform speed $v$. The angle described by a particle in one second is

If the linear speed of a particle moving in a circular path is constant,then the angular speed is proportional to (where $r$ is the radius of the circle):

If $\omega_1$ is the angular velocity of the hour hand of a clock and $\omega_2$ is the angular velocity of the Earth,then the ratio $\omega_1 : \omega_2$ is