$A$ stone is tied to one end of a string $50\, cm$ long and is whirled in a horizontal circle with a constant speed. If the stone makes $10$ revolutions in $20\, s$,what is the magnitude of the acceleration of the stone in $cm/s^2$?

$A$ 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 $\alpha=$ . . . . . . .

$A$ 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 $.0$)

The net applied force on a body in uniform circular motion should always be

$A$ particle is in uniform circular motion. The equation of its trajectory is given by $(x-2)^2+y^2=25$,where $x$ and $y$ are in meters. The speed of the particle is $2 \text{ m/s}$. When the particle attains the lowest $y$ coordinate,the acceleration of the particle is (in $\text{m/s}^2$):

$A$ body is moving on a circle of radius $80 \, m$ with a speed $20 \, m/s$ which is decreasing at the rate of $5 \, m/s^2$ at an instant. The angle made by its acceleration with its velocity is .......... (in $^{\circ}$)