$A$ particle moves with constant angular velocity in a circular path of a certain radius and is acted upon by a certain centripetal force $F$. If the angular velocity is doubled,keeping the radius the same,the new force will be

$A$ car is moving with high velocity when it takes a turn. $A$ force acts on it outwardly because of

$A$ hemispherical bowl of radius $R$ is rotated about its axis of symmetry,which is kept vertical,with angular velocity $\omega$. $A$ small block is kept in the bowl. It remains stationary relative to the bowl surface at a position where the radius makes an angle $\theta$ with the vertical. The friction is absent. The value of $\theta$ is

$A$ car is moving with velocity $v$ at the top of a semi-circular hill of radius $40 \,m$ such that the normal force on it is zero. Find the velocity $(v)$ of the car. [Use $g=10 \,ms^{-2}$] (in $\,ms^{-1}$)

$A$ proton of mass $1.6 \times 10^{-27} \, kg$ moves in a circular orbit of radius $0.10 \, m$ under a centripetal force of $4 \times 10^{-13} \, N$. The frequency of revolution of the proton is about:

One end of a string of length $l$ is connected to a particle of mass $m$ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$,the net force on the particle (directed towards the center) will be ($T$ represents the tension in the string).