If the radius of curvature of the path of two particles of same masses are in the ratio $1 : 2$,then in order to have constant centripetal force,their velocity,should be in the ratio of

$A$ motorcycle driver doubles their velocity when taking a turn. The force exerted outwardly will be

In a conical pendulum,the bob of mass $m$ moves in a horizontal circle of radius $r$ with uniform speed $V$. The string of length $L$ describes a cone of semi-vertical angle $\theta$. The centripetal force acting on the bob is $(g = \text{acceleration due to gravity})$

$A$ car of mass $m$ passes over the top of a convex bridge of radius of curvature $r$ with a velocity $v$. What is the normal force exerted by the bridge on the car?

Two stones of masses $m$ and $3m$ are whirled in horizontal circles,the heavier one in a radius of $(r/3)$ and the lighter one in a radius of $r$. The tangential speed of the lighter stone is $n$ times that of the heavier stone when they experience the same centripetal force. The value of $n$ is:

$A$ particle starts from rest and performs circular motion of constant radius with speed given by $v = \alpha \sqrt{x}$,where $\alpha$ is a constant and $x$ is the distance covered. The correct graph of the magnitude of its tangential acceleration $(a_t)$ and centripetal acceleration $(a_c)$ versus $t$ will be: