The ratio of the angular velocity of the hour hand of a watch to the angular velocity of the rotation of the Earth is:

$A$ particle is performing uniform circular motion. The velocity of the particle at an instant of time is $(6 \hat{i} - 2 \hat{j}) \ m/s$ and the angular velocity is $\vec{\omega} = (a \hat{i} + b \hat{j}) \ rad/s$. Then the value of $\frac{b}{a}$ is:

$A$ particle $P$ is moving in a circle of radius $a$ with a uniform speed $v$. $C$ is the centre of the circle and $AB$ is a diameter. When passing through $B$,the angular velocity of $P$ about $A$ and $C$ are in the ratio:

The length of the second's hand in a watch is $1 \, cm$. The change in velocity of its tip in $15 \, seconds$ is

$A$ particle $P$ moves in a circle of radius $a$ with a constant speed $v$. $C$ is the center of the circle,and $AB$ is a diameter. When the particle passes through point $B$,what is the ratio of its angular velocity with respect to $A$ and $C$ respectively?

The angular velocity of the minute hand of a clock in degrees per second is