If the radius of curvature of the path of two particles of same mass are in the ratio $3:4,$ then in order to have constant centripetal force,their velocities will be in the ratio of:

$A$ uniform circular disc of radius $50\, cm$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0\, rad \,s^{-2}$. Its net acceleration in $m\,s^{-2}$ at the end of $2.0\,s$ is approximately (in $.0$)

$A$ particle is moving in a circle of radius $50 \ cm$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t=0$ is $4 \ m/s$,the time taken to complete the first revolution will be $\frac{1}{\alpha}[1-e^{-2 \pi}] \ s$,where $\alpha=$ . . . . . . .

In the given figure,$a = 15 \, m s^{-2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5 \, m$ at a given instant of time. The speed of the particle is ........ $m/s$.

$A$ particle is confined to rotate in a circular path with decreasing linear speed. Which of the following is correct?

$A$ car is moving with a speed of $30 \,ms^{-1}$ on a circular path of radius $500 \,m$. If its speed is increasing at the rate of $2 \,ms^{-2}$, then find its acceleration. (in $\,ms^{-2}$)